计算几何的模板（大神整理）

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计算几何模板

1.3 多边形

#include <stdlib.h>#include <math.h>#define MAXN 1000#define offset 10000#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)#define _sign(x) ((x)>eps?1:((x)<-eps?2:0))struct point{double x,y;};struct line{point a,b;};double xmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}//判定凸多边形,顶点按顺时针或逆时针给出,允许相邻边共线int is_convex(int n,point* p){int i,s[3]={1,1,1};for (i=0;i<n&&s[1]|s[2];i++)s[_sign(xmult(p[(i+1)%n],p[(i+2)%n],p[i]))]=0;return s[1]|s[2];}//判定凸多边形,顶点按顺时针或逆时针给出,不允许相邻边共线int is_convex_v2(int n,point* p){int i,s[3]={1,1,1};for (i=0;i<n&&s[0]&&s[1]|s[2];i++)s[_sign(xmult(p[(i+1)%n],p[(i+2)%n],p[i]))]=0;return s[0]&&s[1]|s[2];}//判点在凸多边形内或多边形边上,顶点按顺时针或逆时针给出int inside_convex(point q,int n,point* p){int i,s[3]={1,1,1};for (i=0;i<n&&s[1]|s[2];i++)s[_sign(xmult(p[(i+1)%n],q,p[i]))]=0;return s[1]|s[2];}//判点在凸多边形内,顶点按顺时针或逆时针给出,在多边形边上返回0int inside_convex_v2(point q,int n,point* p){int i,s[3]={1,1,1};for (i=0;i<n&&s[0]&&s[1]|s[2];i++)s[_sign(xmult(p[(i+1)%n],q,p[i]))]=0;return s[0]&&s[1]|s[2];}//判点在任意多边形内,顶点按顺时针或逆时针给出//on_edge表示点在多边形边上时的返回值,offset为多边形坐标上限int inside_polygon(point q,int n,point* p,int on_edge=1){point q2;int i=0,count;while (i<n)for (count=i=0,q2.x=rand()+offset,q2.y=rand()+offset;i<n;i++)if (zero(xmult(q,p[i],p[(i+1)%n]))&&(p[i].x-q.x)*(p[(i+1)%n].x-q.x)<eps&&(p[i].y-q.y)*(p[(i+1)%n].y-q.y)<eps)return on_edge;else if (zero(xmult(q,q2,p[i])))break;else if (xmult(q,p[i],q2)*xmult(q,p[(i+1)%n],q2)<-eps&&xmult(p[i],q,p[(i+1)%n])*xmult(p[i],q2,p[(i+1)%n])<-eps)count++;return count&1;}inline int opposite_side(point p1,point p2,point l1,point l2){return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps;}inline int dot_online_in(point p,point l1,point l2){return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;}//判线段在任意多边形内,顶点按顺时针或逆时针给出,与边界相交返回1int inside_polygon(point l1,point l2,int n,point* p){point t[MAXN],tt;int i,j,k=0;if (!inside_polygon(l1,n,p)||!inside_polygon(l2,n,p))return 0;for (i=0;i<n;i++)if (opposite_side(l1,l2,p[i],p[(i+1)%n])&&opposite_side(p[i],p[(i+1)%n],l1,l2))return 0;else if (dot_online_in(l1,p[i],p[(i+1)%n]))t[k++]=l1;else if (dot_online_in(l2,p[i],p[(i+1)%n]))t[k++]=l2;else if (dot_online_in(p[i],l1,l2))t[k++]=p[i];for (i=0;i<k;i++)for (j=i+1;j<k;j++){tt.x=(t[i].x+t[j].x)/2;tt.y=(t[i].y+t[j].y)/2;if (!inside_polygon(tt,n,p))return 0;}return 1;}point intersection(line u,line v){point ret=u.a;double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));ret.x+=(u.b.x-u.a.x)*t;ret.y+=(u.b.y-u.a.y)*t;return ret;}point barycenter(point a,point b,point c){line u,v;u.a.x=(a.x+b.x)/2;u.a.y=(a.y+b.y)/2;u.b=c;v.a.x=(a.x+c.x)/2;v.a.y=(a.y+c.y)/2;v.b=b;return intersection(u,v);}//多边形重心point barycenter(int n,point* p){point ret,t;double t1=0,t2;int i;ret.x=ret.y=0;for (i=1;i<n-1;i++)if (fabs(t2=xmult(p[0],p[i],p[i+1]))>eps){t=barycenter(p[0],p[i],p[i+1]);ret.x+=t.x*t2;ret.y+=t.y*t2;t1+=t2;}if (fabs(t1)>eps)ret.x/=t1,ret.y/=t1;return ret;}

1.4多边形切割

//多边形切割//可用于半平面交#define MAXN 100#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)struct point{double x,y;};double xmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}int same_side(point p1,point p2,point l1,point l2){return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;}point intersection(point u1,point u2,point v1,point v2){point ret=u1;double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));ret.x+=(u2.x-u1.x)*t;ret.y+=(u2.y-u1.y)*t;return ret;}//将多边形沿l1,l2确定的直线切割在side侧切割,保证l1,l2,side不共线void polygon_cut(int& n,point* p,point l1,point l2,point side){point pp[MAXN];int m=0,i;for (i=0;i<n;i++){if (same_side(p[i],side,l1,l2))pp[m++]=p[i];if (!same_side(p[i],p[(i+1)%n],l1,l2)&&!(zero(xmult(p[i],l1,l2))&&zero(xmult(p[(i+1)%n],l1,l2))))pp[m++]=intersection(p[i],p[(i+1)%n],l1,l2);}for (n=i=0;i<m;i++)if (!i||!zero(pp[i].x-pp[i-1].x)||!zero(pp[i].y-pp[i-1].y))p[n++]=pp[i];if (zero(p[n-1].x-p[0].x)&&zero(p[n-1].y-p[0].y))n--;if (n<3)n=0;}

1.5 浮点函数

//浮点几何函数库#include <math.h>#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)struct point{double x,y;};struct line{point a,b;};//计算cross product (P1-P0)x(P2-P0)double xmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double xmult(double x1,double y1,double x2,double y2,double x0,double y0){return (x1-x0)*(y2-y0)-(x2-x0)*(y1-y0);}//计算dot product (P1-P0).(P2-P0)double dmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);}double dmult(double x1,double y1,double x2,double y2,double x0,double y0){return (x1-x0)*(x2-x0)+(y1-y0)*(y2-y0);}//两点距离double distance(point p1,point p2){return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}double distance(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}//判三点共线int dots_inline(point p1,point p2,point p3){return zero(xmult(p1,p2,p3));}int dots_inline(double x1,double y1,double x2,double y2,double x3,double y3){return zero(xmult(x1,y1,x2,y2,x3,y3));}//判点是否在线段上,包括端点int dot_online_in(point p,line l){return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps;}int dot_online_in(point p,point l1,point l2){return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps;}int dot_online_in(double x,double y,double x1,double y1,double x2,double y2){return zero(xmult(x,y,x1,y1,x2,y2))&&(x1-x)*(x2-x)<eps&&(y1-y)*(y2-y)<eps;}//判点是否在线段上,不包括端点int dot_online_ex(point p,line l){return dot_online_in(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y))&&(!zero(p.x-l.b.x)||!zero(p.y-l.b.y));}int dot_online_ex(point p,point l1,point l2){return dot_online_in(p,l1,l2)&&(!zero(p.x-l1.x)||!zero(p.y-l1.y))&&(!zero(p.x-l2.x)||!zero(p.y-l2.y));}int dot_online_ex(double x,double y,double x1,double y1,double x2,double y2){return dot_online_in(x,y,x1,y1,x2,y2)&&(!zero(x-x1)||!zero(y-y1))&&(!zero(x-x2)||!zero(y-y2));}//判两点在线段同侧,点在线段上返回0int same_side(point p1,point p2,line l){return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps;}int same_side(point p1,point p2,point l1,point l2){return xmult(l1,p1,l2)*xmult(l1,p2,l2)>eps;}//判两点在线段异侧,点在线段上返回0int opposite_side(point p1,point p2,line l){return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)<-eps;}int opposite_side(point p1,point p2,point l1,point l2){return xmult(l1,p1,l2)*xmult(l1,p2,l2)<-eps;}//判两直线平行int parallel(line u,line v){return zero((u.a.x-u.b.x)*(v.a.y-v.b.y)-(v.a.x-v.b.x)*(u.a.y-u.b.y));}int parallel(point u1,point u2,point v1,point v2){return zero((u1.x-u2.x)*(v1.y-v2.y)-(v1.x-v2.x)*(u1.y-u2.y));}//判两直线垂直int perpendicular(line u,line v){return zero((u.a.x-u.b.x)*(v.a.x-v.b.x)+(u.a.y-u.b.y)*(v.a.y-v.b.y));}int perpendicular(point u1,point u2,point v1,point v2){return zero((u1.x-u2.x)*(v1.x-v2.x)+(u1.y-u2.y)*(v1.y-v2.y));}//判两线段相交,包括端点和部分重合int intersect_in(line u,line v){if (!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))return !same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);}int intersect_in(point u1,point u2,point v1,point v2){if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2))return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2);return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u2);}//判两线段相交,不包括端点和部分重合int intersect_ex(line u,line v){return opposite_side(u.a,u.b,v)&&opposite_side(v.a,v.b,u);}int intersect_ex(point u1,point u2,point v1,point v2){return opposite_side(u1,u2,v1,v2)&&opposite_side(v1,v2,u1,u2);}//计算两直线交点,注意事先判断直线是否平行!//线段交点请另外判线段相交(同时还是要判断是否平行!)point intersection(line u,line v){point ret=u.a;double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));ret.x+=(u.b.x-u.a.x)*t;ret.y+=(u.b.y-u.a.y)*t;return ret;}point intersection(point u1,point u2,point v1,point v2){point ret=u1;double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));ret.x+=(u2.x-u1.x)*t;ret.y+=(u2.y-u1.y)*t;return ret;}//点到直线上的最近点point ptoline(point p,line l){point t=p;t.x+=l.a.y-l.b.y,t.y+=l.b.x-l.a.x;return intersection(p,t,l.a,l.b);}point ptoline(point p,point l1,point l2){point t=p;t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;return intersection(p,t,l1,l2);}//点到直线距离double disptoline(point p,line l){return fabs(xmult(p,l.a,l.b))/distance(l.a,l.b);}double disptoline(point p,point l1,point l2){return fabs(xmult(p,l1,l2))/distance(l1,l2);}double disptoline(double x,double y,double x1,double y1,double x2,double y2){return fabs(xmult(x,y,x1,y1,x2,y2))/distance(x1,y1,x2,y2);}//点到线段上的最近点point ptoseg(point p,line l){point t=p;t.x+=l.a.y-l.b.y,t.y+=l.b.x-l.a.x;if (xmult(l.a,t,p)*xmult(l.b,t,p)>eps)return distance(p,l.a)<distance(p,l.b)?l.a:l.b;return intersection(p,t,l.a,l.b);}point ptoseg(point p,point l1,point l2){point t=p;t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;if (xmult(l1,t,p)*xmult(l2,t,p)>eps)return distance(p,l1)<distance(p,l2)?l1:l2;return intersection(p,t,l1,l2);}//点到线段距离double disptoseg(point p,line l){point t=p;t.x+=l.a.y-l.b.y,t.y+=l.b.x-l.a.x;if (xmult(l.a,t,p)*xmult(l.b,t,p)>eps)return distance(p,l.a)<distance(p,l.b)?distance(p,l.a):distance(p,l.b);return fabs(xmult(p,l.a,l.b))/distance(l.a,l.b);}double disptoseg(point p,point l1,point l2){point t=p;t.x+=l1.y-l2.y,t.y+=l2.x-l1.x;if (xmult(l1,t,p)*xmult(l2,t,p)>eps)return distance(p,l1)<distance(p,l2)?distance(p,l1):distance(p,l2);return fabs(xmult(p,l1,l2))/distance(l1,l2);}//矢量V以P为顶点逆时针旋转angle并放大scale倍point rotate(point v,point p,double angle,double scale){point ret=p;v.x-=p.x,v.y-=p.y;p.x=scale*cos(angle);p.y=scale*sin(angle);ret.x+=v.x*p.x-v.y*p.y;ret.y+=v.x*p.y+v.y*p.x;return ret;}//p点关于直线L的对称点ponit symmetricalPointofLine(point p, line L){    point p2;    double d;    d = L.a * L.a + L.b * L.b;    p2.x = (L.b * L.b * p.x - L.a * L.a * p.x -             2 * L.a * L.b * p.y - 2 * L.a * L.c) / d;    p2.y = (L.a * L.a * p.y - L.b * L.b * p.y -             2 * L.a * L.b * p.x - 2 * L.b * L.c) / d;    return p2;}//求两点的平分线line bisector(point& a, point& b) {line ab, ans;  ab.set(a, b);double midx = (a.x + b.x)/2.0,midy = (a.y + b.y)/2.0;ans.a = -ab.b, ans.b = -ab.a, ans.c = -ab.b * midx + ab.a * midy;return ans;}// 已知入射线、镜面，求反射线。 // a1,b1,c1为镜面直线方程(a1 x + b1 y + c1 = 0 ,下同)系数;  a2,b2,c2为入射光直线方程系数;  a,b,c为反射光直线方程系数. // 光是有方向的，使用时注意：入射光向量:<-b2,a2>；反射光向量:<b,-a>. // 不要忘记结果中可能会有"negative zeros" void reflect(double a1,double b1,double c1,double a2,double b2,double c2,double &a,double &b,double &c) { double n,m; double tpb,tpa; tpb=b1*b2+a1*a2; tpa=a2*b1-a1*b2; m=(tpb*b1+tpa*a1)/(b1*b1+a1*a1); n=(tpa*b1-tpb*a1)/(b1*b1+a1*a1); if(fabs(a1*b2-a2*b1)<1e-20) { a=a2;b=b2;c=c2; return; } double xx,yy; //(xx,yy)是入射线与镜面的交点。 xx=(b1*c2-b2*c1)/(a1*b2-a2*b1); yy=(a2*c1-a1*c2)/(a1*b2-a2*b1); a=n; b=-m; c=m*yy-xx*n; }

1.6 面积

#include <math.h>struct point{double x,y;};//计算cross product (P1-P0)x(P2-P0)double xmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double xmult(double x1,double y1,double x2,double y2,double x0,double y0){return (x1-x0)*(y2-y0)-(x2-x0)*(y1-y0);}//计算三角形面积,输入三顶点double area_triangle(point p1,point p2,point p3){return fabs(xmult(p1,p2,p3))/2;}double area_triangle(double x1,double y1,double x2,double y2,double x3,double y3){return fabs(xmult(x1,y1,x2,y2,x3,y3))/2;}//计算三角形面积,输入三边长double area_triangle(double a,double b,double c){double s=(a+b+c)/2;return sqrt(s*(s-a)*(s-b)*(s-c));}//计算多边形面积,顶点按顺时针或逆时针给出double area_polygon(int n,point* p){double s1=0,s2=0;int i;for (i=0;i<n;i++)s1+=p[(i+1)%n].y*p[i].x,s2+=p[(i+1)%n].y*p[(i+2)%n].x;return fabs(s1-s2)/2;}

1.7球面

#include <math.h>const double pi=acos(-1);//计算圆心角lat表示纬度,-90<=w<=90,lng表示经度//返回两点所在大圆劣弧对应圆心角,0<=angle<=pidouble angle(double lng1,double lat1,double lng2,double lat2){double dlng=fabs(lng1-lng2)*pi/180;while (dlng>=pi+pi)dlng-=pi+pi;if (dlng>pi)dlng=pi+pi-dlng;lat1*=pi/180,lat2*=pi/180;return acos(cos(lat1)*cos(lat2)*cos(dlng)+sin(lat1)*sin(lat2));}//计算距离,r为球半径double line_dist(double r,double lng1,double lat1,double lng2,double lat2){double dlng=fabs(lng1-lng2)*pi/180;while (dlng>=pi+pi)dlng-=pi+pi;if (dlng>pi)dlng=pi+pi-dlng;lat1*=pi/180,lat2*=pi/180;return r*sqrt(2-2*(cos(lat1)*cos(lat2)*cos(dlng)+sin(lat1)*sin(lat2)));}//计算球面距离,r为球半径inline double sphere_dist(double r,double lng1,double lat1,double lng2,double lat2){return r*angle(lng1,lat1,lng2,lat2);}//球面反射//SGU110// http://acm.sgu.ru/problem.php?contest=0&problem=110#include <cstdio>#include <cmath>const int size = 555;const double eps = 1e-9;struct point {double x, y, z;} centre = {0, 0, 0};struct circle {point o; double r;} cir[size];struct ray {point s, dir;} l;int n;int dcmp (double x){return x < -eps ? -1 : x > eps;}double sqr (double x){return x*x;}double dot (point a, point b){return a.x * b.x + a.y * b.y + a.z * b.z;}double dis2 (point a, point b){return sqr(a.x-b.x) + sqr(a.y-b.y) + sqr(a.z-b.z);}double disToLine2 (point a, ray l){   /**** 点到直线L的距离的平方 **/point tmp;tmp.x =  l.dir.y * (a.z - l.s.z) - l.dir.z * (a.y - l.s.y);tmp.y = -l.dir.x * (a.z - l.s.z) + l.dir.z * (a.x - l.s.x);tmp.z =  l.dir.x * (a.y - l.s.y) - l.dir.y * (a.x - l.s.x); return dis2 (tmp, centre) / dis2 (l.dir, centre);}/**　用解方程(点到圆心的距离为r)法求交点  (下面有向量法求交点, 两者取其一, 都OK)*//* 是向量分量表示发的系数, 必须在射线上,故K必须为正, t是交点***//*bool find (circle p, ray l, double &k, point &t) {double x = l.s.x - p.o.x, y = l.s.y - p.o.y, z = l.s.z - p.o.z;double a = sqr(l.dir.x) + sqr(l.dir.y) + sqr(l.dir.z);double b = 2 * (x*l.dir.x + y*l.dir.y + z*l.dir.z);double c = x*x + y*y + z*z - p.r*p.r;double det = b*b - 4*a*c;//printf ("a = %lf, b = %lf, c = %lf", a, b, c);//printf ("det = %lf\n", det);if (dcmp(det) == -1) return false;k = (-b - sqrt(det)) / a / 2;if (dcmp(k) != 1) return false;t.x = l.s.x + k * l.dir.x;t.y = l.s.y + k * l.dir.y;t.z = l.s.z + k * l.dir.z;return true;}*//**** 用向量法求交点  ***/bool find (circle p, ray l, double &k, point &t){double h2 = disToLine2 (p.o, l);//printf ("h2 = %lf\n", h2);if (dcmp(p.r*p.r - h2) < 0) return false;point tmp;tmp.x = p.o.x - l.s.x;tmp.y = p.o.y - l.s.y;tmp.z = p.o.z - l.s.z;if (dcmp(dot(tmp, l.dir)) <= 0) return false;k = sqrt(dis2(p.o, l.s) - h2) - sqrt(p.r*p.r - h2);double k1 = k / sqrt(dis2(l.dir, centre));t.x = l.s.x + k1 * l.dir.x;t.y = l.s.y + k1 * l.dir.y;t.z = l.s.z + k1 * l.dir.z;return true; }/*计算新射线的起点和方向 */void newRay (ray &l, ray l1, point inter){double k = - 2 * dot(l.dir, l1.dir);l.dir.x += l1.dir.x * k;l.dir.y += l1.dir.y * k;l.dir.z += l1.dir.z * k;l.s = inter;}/* 返回的是最先相交的球的编号,均不相交,返回-1 */int update (){int sign = -1, i;double k = 1e100, tmp;point inter, t;for (i = 1; i <= n; i++){ //找到最先相交的球if (!find (cir[i], l, tmp, t)) continue;if (dcmp (tmp - k) < 0) k = tmp, inter = t, sign = i;}//ray 变向if (sign == -1) return sign;ray l1;l1.s = cir[sign].o;l1.dir.x = (inter.x - l1.s.x) / cir[sign].r;l1.dir.y = (inter.y - l1.s.y) / cir[sign].r;l1.dir.z = (inter.z - l1.s.z) / cir[sign].r;newRay (l, l1, inter);return sign;}int main (){//  freopen ("in", "r", stdin);int i;scanf ("%d", &n);for (i = 1; i <= n; i++) //输入空间的球位置scanf ("%lf%lf%lf%lf", &cir[i].o.x, &cir[i].o.y, &cir[i].o.z, &cir[i].r);scanf ("%lf%lf%lf%lf%lf%lf", &l.s.x, &l.s.y, &l.s.z, &l.dir.x, &l.dir.y, &l.dir.z);for (i = 0; i <= 10; i++){ //最多输出十次相交的球的编号int sign = update ();if (sign == -1) break;if (i == 0) printf ("%d", sign);else if (i < 10) printf (" %d", sign);else printf (" etc.");}puts ("");}

1.8三角形

#include <math.h>struct point{double x,y;};struct line{point a,b;};double distance(point p1,point p2){return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}point intersection(line u,line v){point ret=u.a;double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));ret.x+=(u.b.x-u.a.x)*t;ret.y+=(u.b.y-u.a.y)*t;return ret;}//外心point circumcenter(point a,point b,point c){line u,v;u.a.x=(a.x+b.x)/2;u.a.y=(a.y+b.y)/2;u.b.x=u.a.x-a.y+b.y;u.b.y=u.a.y+a.x-b.x;v.a.x=(a.x+c.x)/2;v.a.y=(a.y+c.y)/2;v.b.x=v.a.x-a.y+c.y;v.b.y=v.a.y+a.x-c.x;return intersection(u,v);}//内心point incenter(point a,point b,point c){line u,v;double m,n;u.a=a;m=atan2(b.y-a.y,b.x-a.x);n=atan2(c.y-a.y,c.x-a.x);u.b.x=u.a.x+cos((m+n)/2);u.b.y=u.a.y+sin((m+n)/2);v.a=b;m=atan2(a.y-b.y,a.x-b.x);n=atan2(c.y-b.y,c.x-b.x);v.b.x=v.a.x+cos((m+n)/2);v.b.y=v.a.y+sin((m+n)/2);return intersection(u,v);}//垂心point perpencenter(point a,point b,point c){line u,v;u.a=c;u.b.x=u.a.x-a.y+b.y;u.b.y=u.a.y+a.x-b.x;v.a=b;v.b.x=v.a.x-a.y+c.y;v.b.y=v.a.y+a.x-c.x;return intersection(u,v);}//重心//到三角形三顶点距离的平方和最小的点//三角形内到三边距离之积最大的点point barycenter(point a,point b,point c){line u,v;u.a.x=(a.x+b.x)/2;u.a.y=(a.y+b.y)/2;u.b=c;v.a.x=(a.x+c.x)/2;v.a.y=(a.y+c.y)/2;v.b=b;return intersection(u,v);}//费马点//到三角形三顶点距离之和最小的点point fermentpoint(point a,point b,point c){point u,v;double step=fabs(a.x)+fabs(a.y)+fabs(b.x)+fabs(b.y)+fabs(c.x)+fabs(c.y);int i,j,k;u.x=(a.x+b.x+c.x)/3;u.y=(a.y+b.y+c.y)/3;while (step>1e-10)for (k=0;k<10;step/=2,k++)for (i=-1;i<=1;i++)for (j=-1;j<=1;j++){v.x=u.x+step*i;v.y=u.y+step*j;if (distance(u,a)+distance(u,b)+distance(u,c)>distance(v,a)+distance(v,b)+distance(v,c))u=v;}return u;}//求曲率半径 三角形内最大可围成面积#include<iostream> #include<cmath> using namespace std; const double pi=3.14159265358979; int main() {    double a,b,c,d,p,s,r,ans,R,x,l; int T=0;while(cin>>a>>b>>c>>d&&a+b+c+d) {T++;l=a+b+c;p=l/2;s=sqrt(p*(p-a)*(p-b)*(p-c));R= s /p;if (d >= l)  ans = s;else if(2*pi*R>=d) ans=d*d/(4*pi);else{r = (l-d)/((l/R)-(2*pi));x = r*r*s/(R*R);ans = s - x + pi * r * r;  }printf("Case %d: %.2lf\n",T,ans); } return 0; }

1.9三维几何

//三维几何函数库#include <math.h>#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)struct point3{double x,y,z;};struct line3{point3 a,b;};struct plane3{point3 a,b,c;};//计算cross product U x Vpoint3 xmult(point3 u,point3 v){point3 ret;ret.x=u.y*v.z-v.y*u.z;ret.y=u.z*v.x-u.x*v.z;ret.z=u.x*v.y-u.y*v.x;return ret;}//计算dot product U . Vdouble dmult(point3 u,point3 v){return u.x*v.x+u.y*v.y+u.z*v.z;}//矢量差 U - Vpoint3 subt(point3 u,point3 v){point3 ret;ret.x=u.x-v.x;ret.y=u.y-v.y;ret.z=u.z-v.z;return ret;}//取平面法向量point3 pvec(plane3 s){return xmult(subt(s.a,s.b),subt(s.b,s.c));}point3 pvec(point3 s1,point3 s2,point3 s3){return xmult(subt(s1,s2),subt(s2,s3));}//两点距离,单参数取向量大小double distance(point3 p1,point3 p2){return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)+(p1.z-p2.z)*(p1.z-p2.z));}//向量大小double vlen(point3 p){return sqrt(p.x*p.x+p.y*p.y+p.z*p.z);}//判三点共线int dots_inline(point3 p1,point3 p2,point3 p3){return vlen(xmult(subt(p1,p2),subt(p2,p3)))<eps;}//判四点共面int dots_onplane(point3 a,point3 b,point3 c,point3 d){return zero(dmult(pvec(a,b,c),subt(d,a)));}//判点是否在线段上,包括端点和共线int dot_online_in(point3 p,line3 l){return zero(vlen(xmult(subt(p,l.a),subt(p,l.b))))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps&&(l.a.z-p.z)*(l.b.z-p.z)<eps;}int dot_online_in(point3 p,point3 l1,point3 l2){return zero(vlen(xmult(subt(p,l1),subt(p,l2))))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps&&(l1.z-p.z)*(l2.z-p.z)<eps;}//判点是否在线段上,不包括端点int dot_online_ex(point3 p,line3 l){return dot_online_in(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y)||!zero(p.z-l.a.z))&&(!zero(p.x-l.b.x)||!zero(p.y-l.b.y)||!zero(p.z-l.b.z));}int dot_online_ex(point3 p,point3 l1,point3 l2){return dot_online_in(p,l1,l2)&&(!zero(p.x-l1.x)||!zero(p.y-l1.y)||!zero(p.z-l1.z))&&(!zero(p.x-l2.x)||!zero(p.y-l2.y)||!zero(p.z-l2.z));}//判点是否在空间三角形上,包括边界,三点共线无意义int dot_inplane_in(point3 p,plane3 s){return zero(vlen(xmult(subt(s.a,s.b),subt(s.a,s.c)))-vlen(xmult(subt(p,s.a),subt(p,s.b)))-vlen(xmult(subt(p,s.b),subt(p,s.c)))-vlen(xmult(subt(p,s.c),subt(p,s.a))));}int dot_inplane_in(point3 p,point3 s1,point3 s2,point3 s3){return zero(vlen(xmult(subt(s1,s2),subt(s1,s3)))-vlen(xmult(subt(p,s1),subt(p,s2)))-vlen(xmult(subt(p,s2),subt(p,s3)))-vlen(xmult(subt(p,s3),subt(p,s1))));}//判点是否在空间三角形上,不包括边界,三点共线无意义int dot_inplane_ex(point3 p,plane3 s){return dot_inplane_in(p,s)&&vlen(xmult(subt(p,s.a),subt(p,s.b)))>eps&&vlen(xmult(subt(p,s.b),subt(p,s.c)))>eps&&vlen(xmult(subt(p,s.c),subt(p,s.a)))>eps;}int dot_inplane_ex(point3 p,point3 s1,point3 s2,point3 s3){return dot_inplane_in(p,s1,s2,s3)&&vlen(xmult(subt(p,s1),subt(p,s2)))>eps&&vlen(xmult(subt(p,s2),subt(p,s3)))>eps&&vlen(xmult(subt(p,s3),subt(p,s1)))>eps;}//判两点在线段同侧,点在线段上返回0,不共面无意义int same_side(point3 p1,point3 p2,line3 l){return dmult(xmult(subt(l.a,l.b),subt(p1,l.b)),xmult(subt(l.a,l.b),subt(p2,l.b)))>eps;}int same_side(point3 p1,point3 p2,point3 l1,point3 l2){return dmult(xmult(subt(l1,l2),subt(p1,l2)),xmult(subt(l1,l2),subt(p2,l2)))>eps;}//判两点在线段异侧,点在线段上返回0,不共面无意义int opposite_side(point3 p1,point3 p2,line3 l){return dmult(xmult(subt(l.a,l.b),subt(p1,l.b)),xmult(subt(l.a,l.b),subt(p2,l.b)))<-eps;}int opposite_side(point3 p1,point3 p2,point3 l1,point3 l2){return dmult(xmult(subt(l1,l2),subt(p1,l2)),xmult(subt(l1,l2),subt(p2,l2)))<-eps;}//判两点在平面同侧,点在平面上返回0int same_side(point3 p1,point3 p2,plane3 s){return dmult(pvec(s),subt(p1,s.a))*dmult(pvec(s),subt(p2,s.a))>eps;}int same_side(point3 p1,point3 p2,point3 s1,point3 s2,point3 s3){return dmult(pvec(s1,s2,s3),subt(p1,s1))*dmult(pvec(s1,s2,s3),subt(p2,s1))>eps;}//判两点在平面异侧,点在平面上返回0int opposite_side(point3 p1,point3 p2,plane3 s){return dmult(pvec(s),subt(p1,s.a))*dmult(pvec(s),subt(p2,s.a))<-eps;}int opposite_side(point3 p1,point3 p2,point3 s1,point3 s2,point3 s3){return dmult(pvec(s1,s2,s3),subt(p1,s1))*dmult(pvec(s1,s2,s3),subt(p2,s1))<-eps;}//判两直线平行int parallel(line3 u,line3 v){return vlen(xmult(subt(u.a,u.b),subt(v.a,v.b)))<eps;}int parallel(point3 u1,point3 u2,point3 v1,point3 v2){return vlen(xmult(subt(u1,u2),subt(v1,v2)))<eps;}//判两平面平行int parallel(plane3 u,plane3 v){return vlen(xmult(pvec(u),pvec(v)))<eps;}int parallel(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3){return vlen(xmult(pvec(u1,u2,u3),pvec(v1,v2,v3)))<eps;}//判直线与平面平行int parallel(line3 l,plane3 s){return zero(dmult(subt(l.a,l.b),pvec(s)));}int parallel(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3){return zero(dmult(subt(l1,l2),pvec(s1,s2,s3)));}//判两直线垂直int perpendicular(line3 u,line3 v){return zero(dmult(subt(u.a,u.b),subt(v.a,v.b)));}int perpendicular(point3 u1,point3 u2,point3 v1,point3 v2){return zero(dmult(subt(u1,u2),subt(v1,v2)));}//判两平面垂直int perpendicular(plane3 u,plane3 v){return zero(dmult(pvec(u),pvec(v)));}int perpendicular(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3){return zero(dmult(pvec(u1,u2,u3),pvec(v1,v2,v3)));}//判直线与平面平行int perpendicular(line3 l,plane3 s){return vlen(xmult(subt(l.a,l.b),pvec(s)))<eps;}int perpendicular(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3){return vlen(xmult(subt(l1,l2),pvec(s1,s2,s3)))<eps;}//判两线段相交,包括端点和部分重合int intersect_in(line3 u,line3 v){if (!dots_onplane(u.a,u.b,v.a,v.b))return 0;if (!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))return !same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);}int intersect_in(point3 u1,point3 u2,point3 v1,point3 v2){if (!dots_onplane(u1,u2,v1,v2))return 0;if (!dots_inline(u1,u2,v1)||!dots_inline(u1,u2,v2))return !same_side(u1,u2,v1,v2)&&!same_side(v1,v2,u1,u2);return dot_online_in(u1,v1,v2)||dot_online_in(u2,v1,v2)||dot_online_in(v1,u1,u2)||dot_online_in(v2,u1,u2);}//判两线段相交,不包括端点和部分重合int intersect_ex(line3 u,line3 v){return dots_onplane(u.a,u.b,v.a,v.b)&&opposite_side(u.a,u.b,v)&&opposite_side(v.a,v.b,u);}int intersect_ex(point3 u1,point3 u2,point3 v1,point3 v2){return dots_onplane(u1,u2,v1,v2)&&opposite_side(u1,u2,v1,v2)&&opposite_side(v1,v2,u1,u2);}//判线段与空间三角形相交,包括交于边界和(部分)包含int intersect_in(line3 l,plane3 s){return !same_side(l.a,l.b,s)&&!same_side(s.a,s.b,l.a,l.b,s.c)&&!same_side(s.b,s.c,l.a,l.b,s.a)&&!same_side(s.c,s.a,l.a,l.b,s.b);}int intersect_in(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3){return !same_side(l1,l2,s1,s2,s3)&&!same_side(s1,s2,l1,l2,s3)&&!same_side(s2,s3,l1,l2,s1)&&!same_side(s3,s1,l1,l2,s2);}//判线段与空间三角形相交,不包括交于边界和(部分)包含int intersect_ex(line3 l,plane3 s){return opposite_side(l.a,l.b,s)&&opposite_side(s.a,s.b,l.a,l.b,s.c)&&opposite_side(s.b,s.c,l.a,l.b,s.a)&&opposite_side(s.c,s.a,l.a,l.b,s.b);}int intersect_ex(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3){return opposite_side(l1,l2,s1,s2,s3)&&opposite_side(s1,s2,l1,l2,s3)&&opposite_side(s2,s3,l1,l2,s1)&&opposite_side(s3,s1,l1,l2,s2);}//计算两直线交点,注意事先判断直线是否共面和平行!//线段交点请另外判线段相交(同时还是要判断是否平行!)point3 intersection(line3 u,line3 v){point3 ret=u.a;double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));ret.x+=(u.b.x-u.a.x)*t;ret.y+=(u.b.y-u.a.y)*t;ret.z+=(u.b.z-u.a.z)*t;return ret;}point3 intersection(point3 u1,point3 u2,point3 v1,point3 v2){point3 ret=u1;double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));ret.x+=(u2.x-u1.x)*t;ret.y+=(u2.y-u1.y)*t;ret.z+=(u2.z-u1.z)*t;return ret;}//计算直线与平面交点,注意事先判断是否平行,并保证三点不共线!//线段和空间三角形交点请另外判断point3 intersection(line3 l,plane3 s){point3 ret=pvec(s);double t=(ret.x*(s.a.x-l.a.x)+ret.y*(s.a.y-l.a.y)+ret.z*(s.a.z-l.a.z))/(ret.x*(l.b.x-l.a.x)+ret.y*(l.b.y-l.a.y)+ret.z*(l.b.z-l.a.z));ret.x=l.a.x+(l.b.x-l.a.x)*t;ret.y=l.a.y+(l.b.y-l.a.y)*t;ret.z=l.a.z+(l.b.z-l.a.z)*t;return ret;}point3 intersection(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3){point3 ret=pvec(s1,s2,s3);double t=(ret.x*(s1.x-l1.x)+ret.y*(s1.y-l1.y)+ret.z*(s1.z-l1.z))/(ret.x*(l2.x-l1.x)+ret.y*(l2.y-l1.y)+ret.z*(l2.z-l1.z));ret.x=l1.x+(l2.x-l1.x)*t;ret.y=l1.y+(l2.y-l1.y)*t;ret.z=l1.z+(l2.z-l1.z)*t;return ret;}//计算两平面交线,注意事先判断是否平行,并保证三点不共线!line3 intersection(plane3 u,plane3 v){line3 ret;ret.a=parallel(v.a,v.b,u.a,u.b,u.c)?intersection(v.b,v.c,u.a,u.b,u.c):intersection(v.a,v.b,u.a,u.b,u.c);ret.b=parallel(v.c,v.a,u.a,u.b,u.c)?intersection(v.b,v.c,u.a,u.b,u.c):intersection(v.c,v.a,u.a,u.b,u.c);return ret;}line3 intersection(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3){line3 ret;ret.a=parallel(v1,v2,u1,u2,u3)?intersection(v2,v3,u1,u2,u3):intersection(v1,v2,u1,u2,u3);ret.b=parallel(v3,v1,u1,u2,u3)?intersection(v2,v3,u1,u2,u3):intersection(v3,v1,u1,u2,u3);return ret;}//点到直线距离double ptoline(point3 p,line3 l){return vlen(xmult(subt(p,l.a),subt(l.b,l.a)))/distance(l.a,l.b);}double ptoline(point3 p,point3 l1,point3 l2){return vlen(xmult(subt(p,l1),subt(l2,l1)))/distance(l1,l2);}//点到平面距离double ptoplane(point3 p,plane3 s){return fabs(dmult(pvec(s),subt(p,s.a)))/vlen(pvec(s));}double ptoplane(point3 p,point3 s1,point3 s2,point3 s3){return fabs(dmult(pvec(s1,s2,s3),subt(p,s1)))/vlen(pvec(s1,s2,s3));}//直线到直线距离double linetoline(line3 u,line3 v){point3 n=xmult(subt(u.a,u.b),subt(v.a,v.b));return fabs(dmult(subt(u.a,v.a),n))/vlen(n);}double linetoline(point3 u1,point3 u2,point3 v1,point3 v2){point3 n=xmult(subt(u1,u2),subt(v1,v2));return fabs(dmult(subt(u1,v1),n))/vlen(n);}//两直线夹角cos值double angle_cos(line3 u,line3 v){return dmult(subt(u.a,u.b),subt(v.a,v.b))/vlen(subt(u.a,u.b))/vlen(subt(v.a,v.b));}double angle_cos(point3 u1,point3 u2,point3 v1,point3 v2){return dmult(subt(u1,u2),subt(v1,v2))/vlen(subt(u1,u2))/vlen(subt(v1,v2));}//两平面夹角cos值double angle_cos(plane3 u,plane3 v){return dmult(pvec(u),pvec(v))/vlen(pvec(u))/vlen(pvec(v));}double angle_cos(point3 u1,point3 u2,point3 u3,point3 v1,point3 v2,point3 v3){return dmult(pvec(u1,u2,u3),pvec(v1,v2,v3))/vlen(pvec(u1,u2,u3))/vlen(pvec(v1,v2,v3));}//直线平面夹角sin值double angle_sin(line3 l,plane3 s){return dmult(subt(l.a,l.b),pvec(s))/vlen(subt(l.a,l.b))/vlen(pvec(s));}double angle_sin(point3 l1,point3 l2,point3 s1,point3 s2,point3 s3){return dmult(subt(l1,l2),pvec(s1,s2,s3))/vlen(subt(l1,l2))/vlen(pvec(s1,s2,s3));}

1.10 凸包

//水平序#define maxn 100005struct point{double x,y;}p[maxn],s[maxn];bool operator < (point a,point b){return a.x<b.x || a.x==b.x&&a.y<b.y;}int n,f;double cp(point a,point b,point c){return (c.y-a.y)*(b.x-a.x)-(b.y-a.y)*(c.x-a.x);}void Convex(point *p,int &n){sort(p,p+n);int i,j,r,top,m;s[0] = p[0];s[1] = p[1];top = 1;for(i=2;i<n;i++){while( top>0 && cp(p[i],s[top],s[top-1])>=0 ) top--;top++;s[top] = p[i];}m = top;top++;s[top] = p[n-2];for(i=n-3;i>=0;i--){while( top>m && cp(p[i],s[top],s[top-1])>=0 ) top--;top++;s[top] = p[i];}top--;n = top+1;}极角序#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>using namespace std;#define maxn 100005int N;struct A{int x,y;int v,l;}P[maxn];int xmult(int x1,int y1,int x2,int y2,int x3,int y3){return (y2-y1)*(x3-x1)-(y3-y1)*(x2-x1);}void swap(A &a,A &b){A t = a;a = b,b = t;}bool operator < (A a,A b){int k = xmult(P[0].x,P[0].y,a.x,a.y,b.x,b.y);if( k<0 )return 1;else if( k==0 ){if( abs(P[0].x-a.x)<abs(P[0].x-b.x) )return 1;if( abs(P[0].y-a.y)<abs(P[0].y-b.y) )return 1;}return 0;}void Grem_scan(int n){int i,j,k,l;k = 0x7fffffff;for(i=0;i<n;i++)if( P[i].x<k || P[i].x==k && P[i].y<P[l].y )k = P[i].x,l = i;swap(P[l],P[0]);sort(P+1,P+n);l = 3;for(i=3;i<n;i++){while( xmult(P[l-2].x,P[l-2].y,P[l-1].x,P[l-1].y,P[i].x,P[i].y)>0 )l--;P[l++] = P[i];}}main(){int i,j,k,l;N = 0;while( scanf("%d%d",&P[N].x,&P[N].y)!=EOF )N++;Grem_scan(N);for(i=0;i<N;i++)if( P[i].x==0 && P[i].y==0 )break;k = i++;printf("(0,0)\n");while( i!=k )printf("(%d,%d)\n",P[i].x,P[i].y),i = (i+1)%N;}//卷包裹法#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define maxn 55struct A{int x,y;}P[maxn];int T,N;bool B[maxn];int as[maxn],L;int xmult(A a,A b,A c){return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);}main(){int i,j,k,l;scanf("%d",&T);while( T-- ){scanf("%d",&N);k = 0x7ffffff;for(i=0;i<N;i++){scanf("%d%d%d",&j,&P[i].x,&P[i].y);if( P[i].y<k )k = P[i].y,l = i;}memset(B,0,sizeof(B));B[l] = 1;as[0] = l;L = 1;while( 1 ){A a,b;if( L==1 )a.x = 0,a.y = P[as[0]].y;elsea = P[as[L-2]];b = P[as[L-1]];k = -1;for(i=0;i<N;i++){if( B[i] )continue;if( xmult(a,b,P[i])<0 )continue;if( k==-1 || xmult(P[as[L-1]],P[k],P[i])<0 || xmult(P[as[L-1]],P[k],P[i])==0 && P[i].y<P[k].y )k = i;}if( k==-1 )break;B[k] = 1;as[L++] = k;}printf("%d ",L);for(i=0;i<L;i++)printf("%d ",as[i]+1);printf("\n");}}}

1.11 网格

#define abs(x) ((x)>0?(x):-(x))struct point{int x,y;};int gcd(int a,int b){return b?gcd(b,a%b):a;}//多边形上的网格点个数int grid_onedge(int n,point* p){int i,ret=0;for (i=0;i<n;i++)ret+=gcd(abs(p[i].x-p[(i+1)%n].x),abs(p[i].y-p[(i+1)%n].y));return ret;}//多边形内的网格点个数int grid_inside(int n,point* p){int i,ret=0;for (i=0;i<n;i++)ret+=p[(i+1)%n].y*(p[i].x-p[(i+2)%n].x);return (abs(ret)-grid_onedge(n,p))/2+1;}

1.12 圆

#include <math.h>#define eps 1e-8struct point{double x,y;};double xmult(point p1,point p2,point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double distance(point p1,point p2){return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}double disptoline(point p,point l1,point l2){return fabs(xmult(p,l1,l2))/distance(l1,l2);}point intersection(point u1,point u2,point v1,point v2){point ret=u1;double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));ret.x+=(u2.x-u1.x)*t;ret.y+=(u2.y-u1.y)*t;return ret;}//判直线和圆相交,包括相切int intersect_line_circle(point c,double r,point l1,point l2){return disptoline(c,l1,l2)<r+eps;}//判线段和圆相交,包括端点和相切int intersect_seg_circle(point c,double r,point l1,point l2){double t1=distance(c,l1)-r,t2=distance(c,l2)-r;point t=c;if (t1<eps||t2<eps)return t1>-eps||t2>-eps;t.x+=l1.y-l2.y;t.y+=l2.x-l1.x;return xmult(l1,c,t)*xmult(l2,c,t)<eps&&disptoline(c,l1,l2)-r<eps;}//判圆和圆相交,包括相切int intersect_circle_circle(point c1,double r1,point c2,double r2){return distance(c1,c2)<r1+r2+eps&&distance(c1,c2)>fabs(r1-r2)-eps;}//计算圆上到点p最近点,如p与圆心重合,返回p本身point dot_to_circle(point c,double r,point p){point u,v;if (distance(p,c)<eps)return p;u.x=c.x+r*fabs(c.x-p.x)/distance(c,p);u.y=c.y+r*fabs(c.y-p.y)/distance(c,p)*((c.x-p.x)*(c.y-p.y)<0?-1:1);v.x=c.x-r*fabs(c.x-p.x)/distance(c,p);v.y=c.y-r*fabs(c.y-p.y)/distance(c,p)*((c.x-p.x)*(c.y-p.y)<0?-1:1);return distance(u,p)<distance(v,p)?u:v;}//计算直线与圆的交点,保证直线与圆有交点//计算线段与圆的交点可用这个函数后判点是否在线段上void intersection_line_circle(point c,double r,point l1,point l2,point& p1,point& p2){point p=c;double t;p.x+=l1.y-l2.y;p.y+=l2.x-l1.x;p=intersection(p,c,l1,l2);t=sqrt(r*r-distance(p,c)*distance(p,c))/distance(l1,l2);p1.x=p.x+(l2.x-l1.x)*t;p1.y=p.y+(l2.y-l1.y)*t;p2.x=p.x-(l2.x-l1.x)*t;p2.y=p.y-(l2.y-l1.y)*t;}//计算圆与圆的交点,保证圆与圆有交点,圆心不重合void intersection_circle_circle(point c1,double r1,point c2,double r2,point& p1,point& p2){point u,v;double t;t=(1+(r1*r1-r2*r2)/distance(c1,c2)/distance(c1,c2))/2;u.x=c1.x+(c2.x-c1.x)*t;u.y=c1.y+(c2.y-c1.y)*t;v.x=u.x+c1.y-c2.y;v.y=u.y-c1.x+c2.x;intersection_line_circle(c1,r1,u,v,p1,p2);}//将向量p逆时针旋转angle角度Point Rotate(Point p,double angle) {    Point res;    res.x=p.x*cos(angle)-p.y*sin(angle);    res.y=p.x*sin(angle)+p.y*cos(angle);    return res;}//求圆外一点对圆(o,r)的两个切点result1和result2void TangentPoint_PC(Point poi,Point o,double r,Point &result1,Point &result2) {    double line=sqrt((poi.x-o.x)*(poi.x-o.x)+(poi.y-o.y)*(poi.y-o.y));    double angle=acos(r/line);    Point unitvector,lin;    lin.x=poi.x-o.x;    lin.y=poi.y-o.y;    unitvector.x=lin.x/sqrt(lin.x*lin.x+lin.y*lin.y)*r;    unitvector.y=lin.y/sqrt(lin.x*lin.x+lin.y*lin.y)*r;    result1=Rotate(unitvector,-angle);    result2=Rotate(unitvector,angle);    result1.x+=o.x;    result1.y+=o.y;    result2.x+=o.x;    result2.y+=o.y;    return;}

1.13 矢量运算求几何模板

#include <iostream>#include <cmath> #include <vector> #include <algorithm> #define MAX_N 100using namespace std; /////////////////////////////////////////////////////////////////////常量区const double INF        = 1e10;     // 无穷大 const double EPS        = 1e-15;    // 计算精度 const int LEFT          = 0;        // 点在直线左边 const int RIGHT         = 1;        // 点在直线右边 const int ONLINE        = 2;        // 点在直线上 const int CROSS         = 0;        // 两直线相交 const int COLINE        = 1;        // 两直线共线 const int PARALLEL      = 2;        // 两直线平行 const int NOTCOPLANAR   = 3;        // 两直线不共面 const int INSIDE        = 1;        // 点在图形内部 const int OUTSIDE       = 2;        // 点在图形外部 const int BORDER        = 3;        // 点在图形边界 const int BAOHAN        = 1;        // 大圆包含小圆const int NEIQIE        = 2;        // 内切const int XIANJIAO      = 3;        // 相交const int WAIQIE        = 4;        // 外切const int XIANLI        = 5;        // 相离const double pi   = acos(-1.0)  //圆周率/////////////////////////////////////////////////////////////////// /////////////////////////////////////////////////////////////////////类型定义区struct Point {              // 二维点或矢量     double x, y;     double angle, dis;     Point() {}     Point(double x0, double y0): x(x0), y(y0) {} }; struct Point3D {            //三维点或矢量     double x, y, z;     Point3D() {}     Point3D(double x0, double y0, double z0): x(x0), y(y0), z(z0) {} }; struct Line {               // 二维的直线或线段     Point p1, p2;     Line() {}     Line(Point p10, Point p20): p1(p10), p2(p20) {} }; struct Line3D {             // 三维的直线或线段     Point3D p1, p2;     Line3D() {}     Line3D(Point3D p10, Point3D p20): p1(p10), p2(p20) {} }; struct Rect {              // 用长宽表示矩形的方法 w, h分别表示宽度和高度     double w, h;  Rect() {} Rect(double _w,double _h) : w(_w),h(_h) {}}; struct Rect_2 {             // 表示矩形，左下角坐标是(xl, yl)，右上角坐标是(xh, yh)     double xl, yl, xh, yh;  Rect_2() {} Rect_2(double _xl,double _yl,double _xh,double _yh) : xl(_xl),yl(_yl),xh(_xh),yh(_yh) {}}; struct Circle {            //圆 Point c; double r; Circle() {} Circle(Point _c,double _r) :c(_c),r(_r) {}};typedef vector<Point> Polygon;      // 二维多边形     typedef vector<Point> Points;       // 二维点集 typedef vector<Point3D> Points3D;   // 三维点集 /////////////////////////////////////////////////////////////////// /////////////////////////////////////////////////////////////////////基本函数区inline double max(double x,double y) {     return x > y ? x : y; } inline double min(double x, double y) {     return x > y ? y : x; } inline bool ZERO(double x)              // x == 0 {     return (fabs(x) < EPS); } inline bool ZERO(Point p)               // p == 0 {     return (ZERO(p.x) && ZERO(p.y)); } inline bool ZERO(Point3D p)              // p == 0 {     return (ZERO(p.x) && ZERO(p.y) && ZERO(p.z)); } inline bool EQ(double x, double y)      // eqaul, x == y {     return (fabs(x - y) < EPS); } inline bool NEQ(double x, double y)     // not equal, x != y {     return (fabs(x - y) >= EPS); } inline bool LT(double x, double y)     // less than, x < y {     return ( NEQ(x, y) && (x < y) ); } inline bool GT(double x, double y)     // greater than, x > y {     return ( NEQ(x, y) && (x > y) ); } inline bool LEQ(double x, double y)     // less equal, x <= y {     return ( EQ(x, y) || (x < y) ); } inline bool GEQ(double x, double y)     // greater equal, x >= y {     return ( EQ(x, y) || (x > y) ); } // 注意！！！ // 如果是一个很小的负的浮点数 // 保留有效位数输出的时候会出现-0.000这样的形式， // 前面多了一个负号 // 这就会导致错误！！！！！！ // 因此在输出浮点数之前，一定要调用次函数进行修正！ inline double FIX(double x) {     return (fabs(x) < EPS) ? 0 : x; } ////////////////////////////////////////////////////////////////////////////////////// ///////////////////////////////////////////////////////////////////////////////////////二维矢量运算 bool operator==(Point p1, Point p2)  {     return ( EQ(p1.x, p2.x) &&  EQ(p1.y, p2.y) ); } bool operator!=(Point p1, Point p2)  {     return ( NEQ(p1.x, p2.x) ||  NEQ(p1.y, p2.y) ); } bool operator<(Point p1, Point p2) {     if (NEQ(p1.x, p2.x)) {         return (p1.x < p2.x);     } else {         return (p1.y < p2.y);     } } Point operator+(Point p1, Point p2)  {     return Point(p1.x + p2.x, p1.y + p2.y); } Point operator-(Point p1, Point p2)  {     return Point(p1.x - p2.x, p1.y - p2.y); } double operator*(Point p1, Point p2) // 计算叉乘 p1 × p2 {     return (p1.x * p2.y - p2.x * p1.y); } double operator&(Point p1, Point p2) { // 计算点积 p1·p2     return (p1.x * p2.x + p1.y * p2.y); } double Norm(Point p) // 计算矢量p的模 {     return sqrt(p.x * p.x + p.y * p.y); } // 把矢量p旋转角度angle (弧度表示) // angle > 0表示逆时针旋转 // angle < 0表示顺时针旋转 Point Rotate(Point p, double angle) {     Point result;     result.x = p.x * cos(angle) - p.y * sin(angle);     result.y = p.x * sin(angle) + p.y * cos(angle);     return result; } ////////////////////////////////////////////////////////////////////////////////////// ////////////////////////////////////////////////////////////////////////////////////// //三维矢量运算 bool operator==(Point3D p1, Point3D p2)  {     return ( EQ(p1.x, p2.x) && EQ(p1.y, p2.y) && EQ(p1.z, p2.z) ); } bool operator<(Point3D p1, Point3D p2) {     if (NEQ(p1.x, p2.x)) {         return (p1.x < p2.x);     } else if (NEQ(p1.y, p2.y)) {         return (p1.y < p2.y);     } else {         return (p1.z < p2.z);     } } Point3D operator+(Point3D p1, Point3D p2)  {     return Point3D(p1.x + p2.x, p1.y + p2.y, p1.z + p2.z); } Point3D operator-(Point3D p1, Point3D p2)  {     return Point3D(p1.x - p2.x, p1.y - p2.y, p1.z - p2.z); } Point3D operator*(Point3D p1, Point3D p2) // 计算叉乘 p1 x p2 {     return Point3D(p1.y * p2.z - p1.z * p2.y,         p1.z * p2.x - p1.x * p2.z,         p1.x * p2.y - p1.y * p2.x );         } double operator&(Point3D p1, Point3D p2) { // 计算点积 p1·p2     return (p1.x * p2.x + p1.y * p2.y + p1.z * p2.z); } double Norm(Point3D p) // 计算矢量p的模 {     return sqrt(p.x * p.x + p.y * p.y + p.z * p.z); } ////////////////////////////////////////////////////////////////////////////////////// ///////////////////////////////////////////////////////////////////////////////////////点.线段.直线问题//double Distance(Point p1, Point p2) //2点间的距离{ return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}double Distance(Point3D p1, Point3D p2) //2点间的距离,三维{ return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)+(p1.z-p2.z)*(p1.z-p2.z));}double Distance(Point p, Line L) // 求二维平面上点到直线的距离 {     return ( fabs((p - L.p1) * (L.p2 - L.p1)) / Norm(L.p2 - L.p1) ); } double Distance(Point3D p, Line3D L)// 求三维空间中点到直线的距离 {     return ( Norm((p - L.p1) * (L.p2 - L.p1)) / Norm(L.p2 - L.p1) ); } bool OnLine(Point p, Line L) // 判断二维平面上点p是否在直线L上 {     return ZERO( (p - L.p1) * (L.p2 - L.p1) ); } bool OnLine(Point3D p, Line3D L) // 判断三维空间中点p是否在直线L上 {     return ZERO( (p - L.p1) * (L.p2 - L.p1) ); } int Relation(Point p, Line L) // 计算点p与直线L的相对关系 ,返回ONLINE,LEFT,RIGHT{     double res = (L.p2 - L.p1) * (p - L.p1);     if (EQ(res, 0)) {         return ONLINE;     } else if (res > 0) {         return LEFT;     } else {         return RIGHT;     } } bool SameSide(Point p1, Point p2, Line L) // 判断点p1, p2是否在直线L的同侧 {     double m1 = (p1 - L.p1) * (L.p2 - L.p1);     double m2 = (p2 - L.p1) * (L.p2 - L.p1);     return GT(m1 * m2, 0); } bool OnLineSeg(Point p, Line L) // 判断二维平面上点p是否在线段l上 {     return ( ZERO( (L.p1 - p) * (L.p2 - p) ) &&         LEQ((p.x - L.p1.x)*(p.x - L.p2.x), 0) &&         LEQ((p.y - L.p1.y)*(p.y - L.p2.y), 0) ); } bool OnLineSeg(Point3D p, Line3D L) // 判断三维空间中点p是否在线段l上 {     return ( ZERO((L.p1 - p) * (L.p2 - p)) &&         EQ( Norm(p - L.p1) + Norm(p - L.p2), Norm(L.p2 - L.p1)) ); } Point SymPoint(Point p, Line L) // 求二维平面上点p关于直线L的对称点 {     Point result;     double a = L.p2.x - L.p1.x;     double b = L.p2.y - L.p1.y;     double t = ( (p.x - L.p1.x) * a + (p.y - L.p1.y) * b ) / (a*a + b*b);     result.x = 2 * L.p1.x + 2 * a * t - p.x;     result.y = 2 * L.p1.y + 2 * b * t - p.y;     return result; } bool Coplanar(Points3D points) // 判断一个点集中的点是否全部共面 {     int i;     Point3D p;     if (points.size() < 4) return true;     p = (points[2] - points[0]) * (points[1] - points[0]);     for (i = 3; i < points.size(); i++) {         if (! ZERO(p & points[i]) ) return false;     }     return true; } bool LineIntersect(Line L1, Line L2) // 判断二维的两直线是否相交 {     return (! ZERO((L1.p1 - L1.p2)*(L2.p1 - L2.p2)) );  // 是否平行 } bool LineIntersect(Line3D L1, Line3D L2) // 判断三维的两直线是否相交 {     Point3D p1 = L1.p1 - L1.p2;     Point3D p2 = L2.p1 - L2.p2;     Point3D p  = p1 * p2;     if (ZERO(p)) return false;      // 是否平行     p = (L2.p1 - L1.p2) * (L1.p1 - L1.p2);     return ZERO(p & L2.p2);         // 是否共面 } bool LineSegIntersect(Line L1, Line L2) // 判断二维的两条线段是否相交 {     return ( GEQ( max(L1.p1.x, L1.p2.x), min(L2.p1.x, L2.p2.x) ) &&         GEQ( max(L2.p1.x, L2.p2.x), min(L1.p1.x, L1.p2.x) ) &&         GEQ( max(L1.p1.y, L1.p2.y), min(L2.p1.y, L2.p2.y) ) &&         GEQ( max(L2.p1.y, L2.p2.y), min(L1.p1.y, L1.p2.y) ) &&         LEQ( ((L2.p1 - L1.p1) * (L1.p2 - L1.p1)) * ((L2.p2 -  L1.p1) * (L1.p2 - L1.p1)), 0 ) &&         LEQ( ((L1.p1 - L2.p1) * (L2.p2 - L2.p1)) * ((L1.p2 -  L2.p1) * (L2.p2 - L2.p1)), 0 ) );              } bool LineSegIntersect(Line3D L1, Line3D L2) // 判断三维的两条线段是否相交 {     // todo     return true; } // 计算两条二维直线的交点，结果在参数P中返回 // 返回值说明了两条直线的位置关系:  COLINE   -- 共线  PARALLEL -- 平行  CROSS    -- 相交 int CalCrossPoint(Line L1, Line L2, Point& P) {     double A1, B1, C1, A2, B2, C2;     A1 = L1.p2.y - L1.p1.y;     B1 = L1.p1.x - L1.p2.x;     C1 = L1.p2.x * L1.p1.y - L1.p1.x * L1.p2.y;     A2 = L2.p2.y - L2.p1.y;     B2 = L2.p1.x - L2.p2.x;     C2 = L2.p2.x * L2.p1.y - L2.p1.x * L2.p2.y;     if (EQ(A1 * B2, B1 * A2))    {         if (EQ( (A1 + B1) * C2, (A2 + B2) * C1 )) {             return COLINE;         } else {             return PARALLEL;         }     } else {         P.x = (B2 * C1 - B1 * C2) / (A2 * B1 - A1 * B2);         P.y = (A1 * C2 - A2 * C1) / (A2 * B1 - A1 * B2);         return CROSS;     } } // 计算两条三维直线的交点，结果在参数P中返回 // 返回值说明了两条直线的位置关系 COLINE   -- 共线  PARALLEL -- 平行  CROSS    -- 相交  NONCOPLANAR -- 不公面 int CalCrossPoint(Line3D L1, Line3D L2, Point3D& P) {     // todo     return 0; } // 计算点P到直线L的最近点 Point NearestPointToLine(Point P, Line L)  {     Point result;     double a, b, t;     a = L.p2.x - L.p1.x;     b = L.p2.y - L.p1.y;     t = ( (P.x - L.p1.x) * a + (P.y - L.p1.y) * b ) / (a * a + b * b);     result.x = L.p1.x + a * t;     result.y = L.p1.y + b * t;     return result; } // 计算点P到线段L的最近点 Point NearestPointToLineSeg(Point P, Line L)  {     Point result;     double a, b, t;     a = L.p2.x - L.p1.x;     b = L.p2.y - L.p1.y;     t = ( (P.x - L.p1.x) * a + (P.y - L.p1.y) * b ) / (a * a + b * b);     if ( GEQ(t, 0) && LEQ(t, 1) ) {         result.x = L.p1.x + a * t;         result.y = L.p1.y + b * t;     } else {         if ( Norm(P - L.p1) < Norm(P - L.p2) ) {             result = L.p1;         } else {             result = L.p2;         }     }     return result; } // 计算险段L1到线段L2的最短距离 double MinDistance(Line L1, Line L2)  {     double d1, d2, d3, d4;     if (LineSegIntersect(L1, L2)) {         return 0;     } else {         d1 = Norm( NearestPointToLineSeg(L1.p1, L2) - L1.p1 );         d2 = Norm( NearestPointToLineSeg(L1.p2, L2) - L1.p2 );         d3 = Norm( NearestPointToLineSeg(L2.p1, L1) - L2.p1 );         d4 = Norm( NearestPointToLineSeg(L2.p2, L1) - L2.p2 );                  return min( min(d1, d2), min(d3, d4) );     } } // 求二维两直线的夹角， // 返回值是0~Pi之间的弧度 double Inclination(Line L1, Line L2) {     Point u = L1.p2 - L1.p1;     Point v = L2.p2 - L2.p1;     return acos( (u & v) / (Norm(u)*Norm(v)) ); } // 求三维两直线的夹角， // 返回值是0~Pi之间的弧度 double Inclination(Line3D L1, Line3D L2) {     Point3D u = L1.p2 - L1.p1;     Point3D v = L2.p2 - L2.p1;     return acos( (u & v) / (Norm(u)*Norm(v)) ); } //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// 判断两个矩形是否相交 // 如果相邻不算相交 bool Intersect(Rect_2 r1, Rect_2 r2) {     return ( max(r1.xl, r2.xl) < min(r1.xh, r2.xh) &&              max(r1.yl, r2.yl) < min(r1.yh, r2.yh) ); } // 判断矩形r2是否可以放置在矩形r1内 // r2可以任意地旋转 //发现原来的给出的方法过不了OJ上的无归之室这题，//所以用了自己的代码bool IsContain(Rect r1, Rect r2)      //矩形的w>h {      if(r1.w >r2.w && r1.h > r2.h) return true;     else     {        double r = sqrt(r2.w*r2.w + r2.h*r2.h) / 2.0;        double alpha = atan2(r2.h,r2.w);        double sita = asin((r1.h/2.0)/r);        double x = r * cos(sita - 2*alpha);        double y = r * sin(sita - 2*alpha);        if(x < r1.w/2.0 && y < r1.h/2.0 && x > 0 && y > -r1.h/2.0) return true;        else return false;     }} //////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////圆Point Center(const Circle & C) //圆心{          return C.c;      }    double Area(const Circle &C){ return pi*C.r*C.r; } double CommonArea(const Circle & A, const Circle & B) //两个圆的公共面积       {          double area = 0.0;          const Circle & M = (A.r > B.r) ? A : B;          const Circle & N = (A.r > B.r) ? B : A;          double D = Distance(Center(M), Center(N));          if ((D < M.r + N.r) && (D > M.r - N.r))          {              double cosM = (M.r * M.r + D * D - N.r * N.r) / (2.0 * M.r * D);              double cosN = (N.r * N.r + D * D - M.r * M.r) / (2.0 * N.r * D);              double alpha = 2.0 * acos(cosM);              double beta  = 2.0 * acos(cosN);              double TM = 0.5 * M.r * M.r * sin(alpha);              double TN = 0.5 * N.r * N.r * sin(beta);              double FM = (alpha / (2*pi)) * Area(M);              double FN = (beta / (2*pi)) * Area(N);              area = FM + FN - TM - TN;          }          else if (D <= M.r - N.r)          {              area = Area(N);          }          return area;      }      bool IsInCircle(const Circle & C, const Rect_2 & rect)//判断圆是否在矩形内(不允许相切){          return (GT(C.c.x - C.r, rect.xl)  &&  LT(C.c.x + C.r, rect.xh)  &&  GT(C.c.y - C.r, rect.yl)  &&  LT(C.c.y + C.r, rect.yh));      }  //判断2圆的位置关系//返回: //BAOHAN   = 1;        // 大圆包含小圆//NEIQIE   = 2;        // 内切//XIANJIAO = 3;        // 相交//WAIQIE   = 4;        // 外切//XIANLI   = 5;        // 相离int CirCir(const Circle &c1, const Circle &c2)//判断2圆的位置关系{ double dis = Distance(c1.c,c2.c); if(LT(dis,fabs(c1.r-c2.r))) return BAOHAN; if(EQ(dis,fabs(c1.r-c2.r))) return NEIQIE; if(LT(dis,c1.r+c2.r) && GT(dis,fabs(c1.r-c2.r))) return XIANJIAO; if(EQ(dis,c1.r+c2.r)) return WAIQIE; return XIANLI;}////////////////////////////////////////////////////////////////////////int main(){ return 0;}

1.14结构体表示几何图形

//计算几何(二维)   #include <cmath>   #include <cstdio>   #include <algorithm>   using namespace std;   typedef double TYPE;   #define Abs(x) (((x)>0)?(x):(-(x)))   #define Sgn(x) (((x)<0)?(-1):(1))   #define Max(a,b) (((a)>(b))?(a):(b))   #define Min(a,b) (((a)<(b))?(a):(b))   #define Epsilon 1e-8   #define Infinity 1e+10   #define PI acos(-1.0)//3.14159265358979323846   TYPE Deg2Rad(TYPE deg){return (deg * PI / 180.0);}   TYPE Rad2Deg(TYPE rad){return (rad * 180.0 / PI);}   TYPE Sin(TYPE deg){return sin(Deg2Rad(deg));}   TYPE Cos(TYPE deg){return cos(Deg2Rad(deg));}   TYPE ArcSin(TYPE val){return Rad2Deg(asin(val));}   TYPE ArcCos(TYPE val){return Rad2Deg(acos(val));}   TYPE Sqrt(TYPE val){return sqrt(val);}  //点   struct POINT   {     TYPE x;     TYPE y;     POINT() : x(0), y(0) {};     POINT(TYPE _x_, TYPE _y_) : x(_x_), y(_y_) {};   };   // 两个点的距离   TYPE Distance(const POINT & a, const POINT & b)   {     return Sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));   }   //线段   struct SEG   {       POINT a; //起点     POINT b; //终点     SEG() {};     SEG(POINT _a_, POINT _b_):a(_a_),b(_b_) {};   };     //直线(两点式)   struct LINE   {     POINT a;     POINT b;     LINE() {};     LINE(POINT _a_, POINT _b_) : a(_a_), b(_b_) {};   };   //直线(一般式)   struct LINE2   {     TYPE A,B,C;     LINE2() {};     LINE2(TYPE _A_, TYPE _B_, TYPE _C_) : A(_A_), B(_B_), C(_C_) {};   };   //两点式化一般式   LINE2 Line2line(const LINE & L) // y=kx+c k=y/x{     LINE2 L2;     L2.A = L.b.y - L.a.y;     L2.B = L.a.x - L.b.x;     L2.C = L.b.x * L.a.y - L.a.x * L.b.y;     return L2;   }   // 引用返回直线 Ax + By + C =0 的系数   void Coefficient(const LINE & L, TYPE & A, TYPE & B, TYPE & C)   {     A = L.b.y - L.a.y;     B = L.a.x - L.b.x;     C = L.b.x * L.a.y - L.a.x * L.b.y;   }   void Coefficient(const POINT & p,const TYPE a,TYPE & A,TYPE & B,TYPE & C)   {     A = Cos(a);     B = Sin(a);     C = - (p.y * B + p.x * A);   }   /判等(值，点，直线)   bool IsEqual(TYPE a, TYPE b)   {     return (Abs(a - b) <Epsilon);   }   bool IsEqual(const POINT & a, const POINT & b)   {     return (IsEqual(a.x, b.x) && IsEqual(a.y, b.y));   }   bool IsEqual(const LINE & A, const LINE & B)   {     TYPE A1, B1, C1;     TYPE A2, B2, C2;     Coefficient(A, A1, B1, C1);     Coefficient(B, A2, B2, C2);     return IsEqual(A1 * B2, A2 * B1) && IsEqual(A1 * C2, A2 * C1) && IsEqual(B1 * C2, B2 * C1);   }   // 矩形   struct RECT   {     POINT a; // 左下点       POINT b; // 右上点       RECT() {};     RECT(const POINT & _a_, const POINT & _b_) { a = _a_; b = _b_; }   };   //矩形化标准   RECT Stdrect(const RECT & q){     TYPE t;     RECT p=q;     if(p.a.x > p.b.x) swap(p.a.x , p.b.x);      if(p.a.y > p.b.y) swap(p.a.y , p.b.y);      return p;   }   //根据下标返回矩形的边     SEG Edge(const RECT & rect, int idx)   {     SEG edge;     while (idx < 0) idx += 4;     switch (idx % 4)     {     case 0: //下边    edge.a = rect.a;       edge.b = POINT(rect.b.x, rect.a.y);       break;     case 1: //右边    edge.a = POINT(rect.b.x, rect.a.y);       edge.b = rect.b;       break;     case 2: //上边      edge.a = rect.b;       edge.b = POINT(rect.a.x, rect.b.y);       break;     case 3: //左边      edge.a = POINT(rect.a.x, rect.b.y);       edge.b = rect.a;       break;     default:       break;     }     return edge;   }   //矩形的面积   TYPE Area(const RECT & rect)   {     return (rect.b.x - rect.a.x) * (rect.b.y - rect.a.y);   }   //两个矩形的公共面积     TYPE CommonArea(const RECT & A, const RECT & B)   {     TYPE area = 0.0;     POINT LL(Max(A.a.x, B.a.x), Max(A.a.y, B.a.y));     POINT UR(Min(A.b.x, B.b.x), Min(A.b.y, B.b.y));     if( (LL.x <= UR.x) && (LL.y <= UR.y) )     {       area = Area(RECT(LL, UR));     }     return area;   }  //判断圆是否在矩形内(不允许相切)   bool IsInCircle(const CIRCLE & circle, const RECT & rect)   {     return (circle.x - circle.r > rect.a.x) &&       (circle.x + circle.r < rect.b.x) &&       (circle.y - circle.r > rect.a.y) &&       (circle.y + circle.r < rect.b.y);   }   //判断矩形是否在圆内(不允许相切)   bool IsInRect(const CIRCLE & circle, const RECT & rect)   {     POINT c,d;     c.x=rect.a.x; c.y=rect.b.y;     d.x=rect.b.x; d.y=rect.a.y;     return (Distance( Center(circle) , rect.a ) < circle.r) &&       (Distance( Center(circle) , rect.b ) < circle.r) &&       (Distance( Center(circle) , c ) < circle.r) &&       (Distance( Center(circle) , d ) < circle.r);   }   //判断矩形是否与圆相离(不允许相切)   bool Isoutside(const CIRCLE & circle, const RECT & rect)   {     POINT c,d;     c.x=rect.a.x; c.y=rect.b.y;     d.x=rect.b.x; d.y=rect.a.y;     return (Distance( Center(circle) , rect.a ) > circle.r) &&       (Distance( Center(circle) , rect.b ) > circle.r) &&       (Distance( Center(circle) , c ) > circle.r) &&       (Distance( Center(circle) , d ) > circle.r) &&       (rect.a.x > circle.x || circle.x > rect.b.x || rect.a.y > circle.y || circle.y > rect.b.y) ||       ((circle.x - circle.r > rect.b.x) ||       (circle.x + circle.r < rect.a.x) ||       (circle.y - circle.r > rect.b.y) ||       (circle.y + circle.r < rect.a.y));   }

1.15四城部分几何模板

/*1.注意实际运用的时候可以用sqrd代替dist提高精度，节省时间*/#include <iostream>#include <math.h>#include <algorithm>using namespace std;const double INF = 10e300;const double EPS = 1e-8;const double PI = acos(-1.0);inline int dblcmp(double a, double b) {if(fabs(a-b) < EPS) return 0;if(a < b) return -1;return 1;}inline double Max(double a, double b) { if(dblcmp(a, b) == 1) return a; return b; }inline double Min(double a, double b) { if(dblcmp(a, b) == 1) return b; return a; }inline double Agl(double deg) { return deg * PI / 180.0; }struct Point { double x, y; void set(double a, double b) { x = a; y = b; } };struct Vec { double x, y; void set(Point& a, Point& b) { x = b.x-a.x; y = b.y-a.y; } };struct Line { double a, b, c; Point st, end;void set(Point& u, Point& v) {a = v.y - u.y; b = u.x - v.x; c = a*u.x + b*u.y; st = u; end = v; } };inline double dist(Point& a, Point& b) { return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); }inline double sqrd(Point& a, Point& b) { return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); }inline double dot(Vec& a, Vec& b) { return a.x * b.x + a.y * b.y; }inline double cross(Vec& a, Vec& b) { return a.x * b.y - a.y * b.x; }inline double cross(Point& a, Point& b, Point& c) {Vec x, y; x.set(a, b); y.set(a, c); return cross(x, y); }//返回1代表a在bc之间 0代表在端点 -1代表在外面inline int between(Point& a, Point& b, Point& c) { Vec x, y; x.set(a,b); y.set(a,c); return dblcmp(dot(x, y),0); }//3维坐标转换 输入是度数void trans(double lat, double log, double& x, double& y, double& z, double radius) {x = radius * cos(lat) * cos(log);y = radius * cos(lat) * sin(log);z = radius * sin(lat);}//求两点的平分线Line bisector(Point& a, Point& b) {Line ab, ans; ab.set(a, b);double midx = (a.x + b.x)/2.0,midy = (a.y + b.y)/2.0;ans.a = -ab.b, ans.b = -ab.a, ans.c = -ab.b * midx + ab.a * midy;return ans;}//线线相交 如果平行 返回-1, 重合返回-2int line_line_intersect(Line& l1, Line& l2, Point& s) {double det = l1.a*l2.b - l2.a*l1.b;    if(dblcmp(det, 0.0) == 0) { //平行或者重合if(dblcmp(point_line_dist(l1.st, l2.st, l2.end, 0), 0) == 0) return -2;return -1;}    s.x = (l2.b*l1.c - l1.b*l2.c)/det;    s.y = (l1.a*l2.c - l2.a*l1.c)/det;return 1;}//2线段相交 ab, cd 交点是s 平行返回-1, 重合返回-2, 不在线段上面返回0 在线段中间返回1 在线段两端返回2int seg_seg_intersect(Point& a, Point& b, Point& c, Point& d, Point& s) {    Line l1, l2; l1.set(a, b); l2.set(c, d);int ans = line_line_intersect(l1, l2, s);if(ans == 1) {if(between(s, a, b) == 1 && between(s, c, d) == 1) return 1;if(between(s, a, b) == -1 && between(s, c, d) == -1)return 0;return 2;}return ans;}//求三点共圆 中心放在center中 返回半径double center_3point(Point& a, Point& b, Point& c, Point& center) {Line x = bisector(a, b), y = bisector(b, c);line_line_intersect(x, y, center);return dist(center, a);}

1.16 一些代码

1.16.1 最小圆覆盖_zju1450

/*
包含点集所有点的最小圆的算法
最小圆覆盖

http://acm.zju.edu.cn/show_problem.php?pid=1450

相关题目最小球包含 http://acm.pku.edu.cn/JudgeOnline/problem?id=2069

平面上有n个点，给定n个点的坐标，试找一个半径最小的圆，将n
个点全部包围，点可以在圆上。

1. 在点集中任取3点A,B,C。
2. 作一个包含A,B,C三点的最小圆,圆周可能通过这3点，也可能只通过
其中两点,但包含第3点.后一种情况圆周上的两点一定是位于圆的一条直
径的两端。
3. 在点集中找出距离第2步所建圆圆心最远的D点，若D点已在圆内或圆周上，
则该圆即为所求的圆，算法结束.则，执行第4步。
4. 在A,B,C,D中选3个点,使由它们生成的一个包含这4个点的圆为最小，这3
点成为新的A,B,C，返回执行第2步。若在第4步生成的圆的圆周只通过A,B,C,D
中的两点，则圆周上的两点取成新的A和B,从另两点中任取一点作为新的C。

程序设计题解上的解题报告：
对于一个给定的点集A，记MinCircle(A)为点集A的最小外接圆，显然，对于所
有的点集情况A,MinCircle(A)都是存在且惟一的。需要特别说明的是，当A为空
集时，MinCircle(A)为空集，当A={a}时，MinCircle(A)圆心坐标为a，半径为0；
显然，MinCircle(A)可以有A边界上最多三个点确定(当点集A中点的个数大于
1时，有可能两个点确定了MinCircle(A))，也就是说存在着一个点集B，|B|<=3
且B包含与A，有MinCircle(B)=MinCircle(A).所以，如果a不属于B，则
MinCircle(A-{a})=MinCircle(A);如果MinCircle(A-{a})不等于MinCircle(A),则
a属于B。
所以我们可以从一个空集R开始，不断的把题目中给定的点集中的点加入R，同
时维护R的外接圆最小，这样就可以得到解决该题的算法。

pku2069

*/#include <stdio.h>#include <math.h>const int maxn = 1005;//const double eps = 1e-6;struct TPoint{double x, y;TPoint operator-(TPoint &a){TPoint p1;p1.x = x - a.x;p1.y = y - a.y;return p1;}};struct TCircle{double r;TPoint centre;};struct TTriangle{TPoint t[3];};TCircle c;TPoint a[maxn];double distance(TPoint p1, TPoint p2){TPoint p3;p3.x = p2.x - p1.x;p3.y = p2.y - p1.y;return sqrt(p3.x * p3.x + p3.y * p3.y);}double triangleArea(TTriangle t){TPoint p1, p2;p1 = t.t[1] - t.t[0];p2 = t.t[2] - t.t[0];return fabs(p1.x * p2.y - p1.y * p2.x) / 2;}TCircle circumcircleOfTriangle(TTriangle t){    //三角形的外接圆    TCircle tmp;    double a, b, c, c1, c2;    double xA, yA, xB, yB, xC, yC;    a = distance(t.t[0], t.t[1]);    b = distance(t.t[1], t.t[2]);    c = distance(t.t[2], t.t[0]);    //根据S = a * b * c / R / 4;求半径R     tmp.r = a * b * c / triangleArea(t) / 4;        xA = t.t[0].x; yA = t.t[0].y;    xB = t.t[1].x; yB = t.t[1].y;    xC = t.t[2].x; yC = t.t[2].y;    c1 = (xA * xA + yA * yA - xB * xB - yB * yB) / 2;    c2 = (xA * xA + yA * yA - xC * xC - yC * yC) / 2;        tmp.centre.x = (c1 * (yA - yC) - c2 * (yA - yB)) /          ((xA - xB) * (yA - yC) - (xA - xC) * (yA - yB));     tmp.centre.y = (c1 * (xA - xC) - c2 * (xA - xB)) /          ((yA - yB) * (xA - xC) - (yA - yC) * (xA - xB));              return tmp;     }TCircle MinCircle2(int tce, TTriangle ce){TCircle tmp;if(tce == 0) tmp.r = -2;else if(tce == 1) {tmp.centre = ce.t[0];tmp.r = 0;}else if(tce == 2){tmp.r = distance(ce.t[0], ce.t[1]) / 2;tmp.centre.x = (ce.t[0].x + ce.t[1].x) / 2;tmp.centre.y = (ce.t[0].y + ce.t[1].y) / 2; }else if(tce == 3) tmp = circumcircleOfTriangle(ce);return tmp;}void MinCircle(int t, int tce, TTriangle ce){int i, j;TPoint tmp;c = MinCircle2(tce, ce);if(tce == 3) return;for(i = 1;i <= t;i++){if(distance(a[i], c.centre) > c.r){ce.t[tce] = a[i];MinCircle(i - 1, tce + 1, ce);tmp = a[i];for(j = i;j >= 2;j--){a[j] = a[j - 1];}a[1] = tmp;}}}void run(int n){TTriangle ce;int i;MinCircle(n, 0, ce);printf("%.2lf %.2lf %.2lf\n", c.centre.x, c.centre.y, c.r);}int main(){    freopen("circle.in", "r", stdin);    freopen("out.txt", "w", stdout);    int n;while(scanf("%d", &n) != EOF && n){for(int i = 1;i <= n;i++)scanf("%lf%lf", &a[i].x, &a[i].y);run(n);}return 0;}

1.16.2 直线旋转_两凸包的最短距离(poj3608)

#include <stdio.h>#include <math.h>#define pi acos(-1.0)#define eps 1e-6#define inf 1e250#define Maxn 10005typedef struct TPoint{double x, y;}TPoint;typedef struct TPolygon{TPoint p[Maxn];int n;}TPolygon;typedef struct TLine{double a, b, c;}TLine;double max(double a, double b){if(a > b) return a;return b;}double min(double a, double b){if(a < b) return a;return b;}double distance(TPoint p1, TPoint p2){return sqrt((p1.x - p2.x) * (p1.x - p2.x)  + (p1.y - p2.y) * (p1.y - p2.y));}TLine lineFromSegment(TPoint p1, TPoint p2){    TLine tmp;    tmp.a = p2.y - p1.y;    tmp.b = p1.x - p2.x;    tmp.c = p2.x * p1.y - p1.x * p2.y;    return tmp;}double polygonArea(TPolygon p){    int i, n;    double area;    n = p.n;    area = 0;    for(i = 1;i <= n;i++)area += (p.p[i - 1].x * p.p[i % n].y - p.p[i % n].x * p.p[i - 1].y);return area / 2;  }void ChangeClockwise(TPolygon &polygon){TPoint tmp;int i;for(i = 0;i <= (polygon.n - 1) / 2;i++){tmp = polygon.p[i];polygon.p[i] = polygon.p[polygon.n - 1 - i];polygon.p[polygon.n - 1 - i] = tmp;}}double disPointToSeg(TPoint p1, TPoint p2, TPoint p3){    double a = distance(p1, p2);    double b = distance(p1, p3);    double c = distance(p2, p3);    if(fabs(a + b - c) < eps) return 0;    if(fabs(a + c - b) < eps || fabs(b + c - a) < eps) return min(a, b);    double t1 = -a * a + b * b + c * c;    double t2 = a * a - b * b + c * c;    if(t1 <= 0 || t2 <= 0) return min(a, b);        TLine l1 = lineFromSegment(p2, p3);    return fabs(l1.a * p1.x + l1.b * p1.y + l1.c) / sqrt(l1.a * l1.a + l1.b * l1.b);      }double disPallSeg(TPoint p1, TPoint p2, TPoint p3, TPoint p4){return min(min(disPointToSeg(p1, p3, p4), disPointToSeg(p2, p3, p4)), min(disPointToSeg(p3, p1, p2), disPointToSeg(p4, p1, p2)));}double angle(TPoint p1, TPoint p2, double SlewRate){double ang, tmp;TPoint p;p.x = p2.x - p1.x;p.y = p2.y - p1.y;if(fabs(p.x) < eps){if(p.y > 0) ang = pi / 2;else ang = 3 * pi / 2;}else {        ang = atan(p.y / p.x);if(p.x < 0) ang += pi;}while(ang < 0) ang += 2 * pi;if(ang >= pi) SlewRate += pi;if(ang > SlewRate) tmp = ang - SlewRate;else tmp = pi - (SlewRate - ang);while(tmp >= pi) tmp -= pi;if(fabs(tmp - pi) < eps) tmp = 0;return tmp;}int main(){int n, m, i;TPolygon polygon1, polygon2;double ymin1, ymax2, ans, d;int k1, k2;while(scanf("%d%d", &n, &m) && n){polygon1.n = n;polygon2.n = m;for(i = 0;i < n;i++)scanf("%lf%lf", &polygon1.p[i].x, &polygon1.p[i].y);for(i = 0;i < m;i++)scanf("%lf%lf", &polygon2.p[i].x, &polygon2.p[i].y);if(polygonArea(polygon1) < 0) ChangeClockwise(polygon1);if(polygonArea(polygon2) < 0) ChangeClockwise(polygon2);ymin1 = inf, ymax2 = -inf;for(i = 0;i < n;i++)if(polygon1.p[i].y < ymin1) ymin1 = polygon1.p[i].y , k1 = i;for(i = 0;i < m;i++)if(polygon2.p[i].y > ymax2) ymax2 = polygon2.p[i].y , k2 = i;double SlewRate = 0;double angle1, angle2;ans = inf;double Slope = 0;while(Slope <= 360){while(SlewRate >= pi) SlewRate -= pi;if(fabs(pi - SlewRate) < eps) SlewRate = 0;angle1 = angle(polygon1.p[k1], polygon1.p[(k1 + 1) % n], SlewRate);angle2 = angle(polygon2.p[k2], polygon2.p[(k2 + 1) % m], SlewRate);if(fabs(angle1 - angle2) < eps){d = disPallSeg(polygon1.p[k1], polygon1.p[(k1 + 1) % n], polygon2.p[k2], polygon2.p[(k2 + 1) % m]); if(d < ans) ans = d;                k1++;k1 %= n;k2++;k2 %= m; SlewRate += angle1;Slope += angle1;}else if(angle1 < angle2){d = disPointToSeg(polygon2.p[k2], polygon1.p[k1], polygon1.p[(k1 + 1) % n]);if(d < ans) ans = d;k1++;k1 %= n;SlewRate += angle1;Slope += angle1;}else {d = disPointToSeg(polygon1.p[k1], polygon2.p[k2], polygon2.p[(k2 + 1) % m]);if(d < ans) ans = d;k2++;k2 %= m;SlewRate += angle2;Slope += angle2;}}printf("%.5lf\n", ans);}return 0;}

1.16.3 扇形的重心

//Xc = 2*R*sinA/3/A //A为圆心角的一半#include <stdio.h>#include <math.h>int main(){double r, angle;while(scanf("%lf%lf", &r, &angle) != EOF){angle /= 2;printf("%.6lf\n", 2 * r * sin(angle) / 3 / angle);}return 0;}

1.16.4 根据经度纬度求球面距离

/*假设地球是球体， 设地球上某点的经度为lambda,纬度为phi， 则这点的空间坐标是 x=cos(phi)*cos(lambda) y=cos(phi)*sin(lambda) z=sin(phi) 设地球上两点的空间坐标分别为(x1,y1,z1),(x2,y2,z2) 直线距离即为R*sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)+(z2-z1)*(z2-z1)),则它们的夹角为 A = acos(x1 * x2 + y1 * y2 + z1 * z2)，则两地距离为 A * R，其中R为地球平均半径6371 *//*这里坐标都要乘以半径R，但由于是求角度，所以统一都没有乘 注意这里还要判断坐标的正负和经度纬度的规定有关 pku_3407*/#include <stdio.h>#include <math.h>const double pi = acos(-1.0);struct TPoint{   double x, y, z;}; int main(){    double w1, wm1, j1, jm1, wd1, wd2;    double w2, wm2, j2, jm2, jd1, jd2;    TPoint p1, p2;    char chr1, chr2;    while(scanf("%lf%lf ", &w1, &wm1) != EOF){        scanf("%c ", &chr1);        scanf("%lf %lf %c", &j1, &jm1, &chr2);        wd1 = (w1 + wm1 / 60) * pi / 180;        jd1 = (j1 + jm1 / 60) * pi / 180;        if(chr1 == 'S') wd1 *= -1.0;        if(chr2 == 'W') jd1 *= -1.0;        p1.x = cos(wd1) * cos(jd1);        p1.y = cos(wd1) * sin(jd1);        p1.z = sin(wd1);        scanf("%lf %lf %c %lf %lf %c", &w2, &wm2, &chr1, &j2, &jm2, &chr2);        wd2 = (w2 + wm2 / 60) * pi / 180;        jd2 = (j2 + jm2 / 60) * pi / 180;        if(chr1 == 'S') wd2 *= -1.0;        if(chr2 == 'W') jd2 *= -1.0;        p2.x = cos(wd2) * cos(jd2);        p2.y = cos(wd2) * sin(jd2);        p2.z = sin(wd2);        double a = acos(p1.x * p2.x + p1.y * p2.y + p1.z * p2.z);        printf("%.3lf\n", a * 6370.0);       }    return 0;}

1.16.5 多边形的重心

/*题目描述：有一个密度均匀的平面N多边形(3 <= N <= 1000000)，可能凹也可能凸，但没有边相交叉，另外已知N个有序(顺时针或逆时针)顶点的坐标值，第j个顶点坐标为（Xi ， Yi ），且满足 (|Xi|, |Yi| <= 20000)，求这个平面多边形的重心。  解题过程：  从第1个顶点出发，分别连接第i, i+1个顶点组成三角形Ti,1 < i < n,   一共n-2个三角形正好是多连形的一个划分，分别求出每个三角形的面积Si，  总面积为各个面积相加  根据物理学知识得：n个点(xi,yi)每个重量是mi,则重心是     X = (x1*M1+x2*M2+...+xn*Mn) / (M1+M2+....+Mn)     Y = (y1*M1+y2*M2+...+yn*Mn) / (M1+M2+....+Mn)     另个需要用的知识有：  已知3点求三角形的面积，设三点分别为p[0].x, p[0].y p[1].x, p[1].y p[1].x, p[1].y  面积s =[ p[0].x*p[1].y-p[1].x*p[0].y + p[1].x*p[2].y-p[2].x*p[1].y + p[2].x*p[0].y-p[0].x*p[2].y ] / 2 ,   这是这3个点是逆时针的值，顺时针取负。  已知3点求重心，x = (p[0].x+p[1].x+p[2].x)/3.0 , y = (p[0].y+p[1].y+p[2].y)/3.0   另外在求解的过程中，不需要考虑点的输入顺序是顺时针还是逆时针，相除后就抵消了，  还要注意 一点是不必在求每个小三角形的重心时都除以3，可以在最后除一下  *//*fzu_1132*/#include <stdio.h>#include <math.h>typedef struct TPoint{    double x;    double y;}TPoint;double triangleArea(TPoint p0, TPoint p1, TPoint p2){    //已知三角形三个顶点的坐标，求三角形的面积     double k = p0.x * p1.y + p1.x * p2.y+ p2.x * p0.y - p1.x * p0.y - p2.x * p1.y - p0.x * p2.y;//if(k >= 0) return k / 2;//else return -k / 2;           return k / 2;}int main(){    int i, n, test;TPoint p0, p1, p2, center;    double area, sumarea, sumx, sumy;        scanf("%d", &test);      while(test--){scanf("%d", &n);scanf("%lf%lf", &p0.x, &p0.y);scanf("%lf%lf", &p1.x, &p1.y);        sumx = 0;        sumy = 0;        sumarea = 0;        for(i = 2;i < n;i++){scanf("%lf%lf", &p2.x, &p2.y);            center.x = p0.x + p1.x + p2.x;            center.y = p0.y + p1.y + p2.y;              area =  triangleArea(p0, p1, p2);            sumarea += area;sumx += center.x * area;sumy += center.y * area; p1 = p2;                   }        printf("%.2lf %.2lf\n", sumx / sumarea / 3, sumy / sumarea / 3);    }    return 0;}

1.16.6 存不存在一个平面把两堆点分开(poj3643)

#include <stdio.h>struct point{double x, y, z;}pa[201], pb[201];int main() { int n, m, i; while (scanf("%d", &n), n != -1) { for (i = 0; i < n; i++) scanf("%lf%lf%lf", &pa[i].x, &pa[i].y, &pa[i].z); scanf("%d", &m); for (i = 0; i < m; i++) scanf("%lf%lf%lf", &pb[i].x, &pb[i].y, &pb[i].z);int cnt = 0, finish = 0; double a = 0, b = 0, c = 0, d = 0; while (cnt < 100000 && !finish){ finish = 1; for (i = 0; i < n; i++) if (a * pa[i].x + b * pa[i].y + c * pa[i].z + d > 0) { a -= pa[i].x; b -= pa[i].y; c -= pa[i].z; d -= 3; finish = 0; }for (i = 0; i < m; i++) if (a * pb[i].x + b * pb[i].y + c * pb[i].z + d <= 0) { a += pb[i].x; b += pb[i].y; c += pb[i].z; d += 3; finish = 0; }cnt++; }printf("%lf %lf %lf %lf\n", a, b, c, d); }return 0;}

1.16.7 pku_3335_判断多边形的核是否存在

/*多边形的核*/#include <stdio.h>#include <math.h>#define Maxn 3005const double eps = 1e-10;typedef struct TPodouble {double x;double y;}TPoint;typedef struct TPolygon{TPoint p[Maxn];int n;};typedef struct TLine{double a, b, c;}TLine;bool same(TPoint p1, TPoint p2){if(p1.x != p2.x) return false;if(p1.y != p2.y) return false;return true;} double multi(TPoint p1, TPoint p2, TPoint p0){    //求矢量[p0, p1], [p0, p2]的叉积    //p0是顶点     return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);    //若结果等于0，则这三点共线    //若结果大于0，则p0p2在p0p1的逆时针方向    //若结果小于0，则p0p2在p0p1的顺时针方向 }TLine lineFromSegment(TPoint p1, TPoint p2){    //线段所在直线,返回直线方程的三个系统     TLine tmp;    tmp.a = p2.y - p1.y;    tmp.b = p1.x - p2.x;    tmp.c = p2.x * p1.y - p1.x * p2.y;    return tmp;}TPoint LineInter(TLine l1, TLine l2){    //求两直线得交点坐标    TPoint tmp;     double a1 = l1.a;    double b1 = l1.b;    double c1 = l1.c;    double a2 = l2.a;    double b2 = l2.b;    double c2 = l2.c;    //注意这里b1 = 0     if(fabs(b1) < eps){        tmp.x = -c1 / a1;          tmp.y = (-c2 - a2 * tmp.x) / b2;    }           else{        tmp.x = (c1 * b2 - b1 * c2) / (b1 * a2 - b2 * a1);        tmp.y = (-c1 - a1 * tmp.x) / b1;    }return tmp;}TPolygon Cut_polygon(TPoint p1, TPoint p2, TPolygon polygon){TPolygon new_polygon;TPoint interp;TLine l1, l2;int i, j;double t1, t2;new_polygon.n = 0;for(i = 0;i <= polygon.n - 1;i++){t1 = multi(p2, polygon.p[i], p1);t2 = multi(p2, polygon.p[i + 1], p1);if(fabs(t1) < eps || fabs(t2) < eps){if(fabs(t1) < eps) new_polygon.p[new_polygon.n++] = polygon.p[i];if(fabs(t2) < eps) new_polygon.p[new_polygon.n++] = polygon.p[i + 1];}else if(t1 < 0 && t2 < 0){new_polygon.p[new_polygon.n++] = polygon.p[i];new_polygon.p[new_polygon.n++] = polygon.p[i + 1];}else if(t1 * t2  < 0){l1 = lineFromSegment(p1, p2);l2 = lineFromSegment(polygon.p[i], polygon.p[i + 1]);interp = LineInter(l1, l2);if(t1 < 0) {new_polygon.p[new_polygon.n++] = polygon.p[i];new_polygon.p[new_polygon.n++] = interp;}else {new_polygon.p[new_polygon.n++] = interp;new_polygon.p[new_polygon.n++] = polygon.p[i + 1];}}}polygon.n = 0;if(new_polygon.n == 0) return polygon;polygon.p[polygon.n++] = new_polygon.p[0];for(i = 1;i < new_polygon.n;i++){if(!same(new_polygon.p[i], new_polygon.p[i - 1])){polygon.p[polygon.n++] = new_polygon.p[i];}}if(polygon.n != 1 && same(polygon.p[polygon.n - 1], polygon.p[0])) polygon.n--;polygon.p[polygon.n] = polygon.p[0];return polygon;}double polygonArea(TPolygon p){    //已知多边形各顶点的坐标，求其面积    int i, n;    double area;    n = p.n;    area = 0;    for(i = 1;i <= n;i++){        area += (p.p[i - 1].x * p.p[i % n].y - p.p[i % n].x * p.p[i - 1].y);    }  return area / 2;  }void ChangeClockwise(TPolygon &polygon){TPoint tmp;int i;for(i = 0;i <= (polygon.n - 1) / 2;i++){tmp = polygon.p[i];polygon.p[i] = polygon.p[polygon.n - 1 - i];polygon.p[polygon.n - 1 - i] = tmp;}}int main(){int test, i, j;double area;TPolygon polygon, new_polygon;scanf("%d", &test);while(test--){scanf("%d", &polygon.n);for(i = 0;i <= polygon.n - 1;i++){scanf("%lf%lf", &polygon.p[i].x, &polygon.p[i].y);}/*若是逆时针转化为顺时针*/if(polygonArea(polygon) > 0) ChangeClockwise(polygon);polygon.p[polygon.n] = polygon.p[0];new_polygon = polygon;for(i = 0;i <= polygon.n - 1;i++){new_polygon = Cut_polygon(polygon.p[i], polygon.p[i + 1], new_polygon);}area = polygonArea(new_polygon);if(area < 0) printf("%.2lf\n", -area);else printf("%.2lf\n", area);}return 0;}//是否存在#include <stdio.h>#include <math.h>#define Maxn 3005const double eps = 1e-10;typedef struct TPodouble {double x;double y;}TPoint;typedef struct TPolygon{TPoint p[Maxn];int n;};typedef struct TLine{double a, b, c;}TLine;bool same(TPoint p1, TPoint p2){if(p1.x != p2.x) return false;if(p1.y != p2.y) return false;return true;} double multi(TPoint p1, TPoint p2, TPoint p0){    //求矢量[p0, p1], [p0, p2]的叉积    //p0是顶点     return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);    //若结果等于0，则这三点共线    //若结果大于0，则p0p2在p0p1的逆时针方向    //若结果小于0，则p0p2在p0p1的顺时针方向 }TLine lineFromSegment(TPoint p1, TPoint p2){    //线段所在直线,返回直线方程的三个系统     TLine tmp;    tmp.a = p2.y - p1.y;    tmp.b = p1.x - p2.x;    tmp.c = p2.x * p1.y - p1.x * p2.y;    return tmp;}TPoint LineInter(TLine l1, TLine l2){    //求两直线得交点坐标    TPoint tmp;     double a1 = l1.a;    double b1 = l1.b;    double c1 = l1.c;    double a2 = l2.a;    double b2 = l2.b;    double c2 = l2.c;    //注意这里b1 = 0     if(fabs(b1) < eps){        tmp.x = -c1 / a1;          tmp.y = (-c2 - a2 * tmp.x) / b2;    }           else{        tmp.x = (c1 * b2 - b1 * c2) / (b1 * a2 - b2 * a1);        tmp.y = (-c1 - a1 * tmp.x) / b1;    }return tmp;}TPolygon Cut_polygon(TPoint p1, TPoint p2, TPolygon polygon){TPolygon new_polygon;TPoint interp;TLine l1, l2;int i, j;double t1, t2;new_polygon.n = 0;for(i = 0;i <= polygon.n - 1;i++){t1 = multi(p2, polygon.p[i], p1);t2 = multi(p2, polygon.p[i + 1], p1);if(fabs(t1) < eps || fabs(t2) < eps){if(fabs(t1) < eps) new_polygon.p[new_polygon.n++] = polygon.p[i];if(fabs(t2) < eps) new_polygon.p[new_polygon.n++] = polygon.p[i + 1];}else if(t1 < 0 && t2 < 0){new_polygon.p[new_polygon.n++] = polygon.p[i];new_polygon.p[new_polygon.n++] = polygon.p[i + 1];}else if(t1 * t2  < 0){l1 = lineFromSegment(p1, p2);l2 = lineFromSegment(polygon.p[i], polygon.p[i + 1]);interp = LineInter(l1, l2);if(t1 < 0) {new_polygon.p[new_polygon.n++] = polygon.p[i];new_polygon.p[new_polygon.n++] = interp;}else {new_polygon.p[new_polygon.n++] = interp;new_polygon.p[new_polygon.n++] = polygon.p[i + 1];}}}polygon.n = 0;if(new_polygon.n == 0) return polygon;polygon.p[polygon.n++] = new_polygon.p[0];for(i = 1;i < new_polygon.n;i++){if(!same(new_polygon.p[i], new_polygon.p[i - 1])){polygon.p[polygon.n++] = new_polygon.p[i];}}if(polygon.n != 1 && same(polygon.p[polygon.n - 1], polygon.p[0])) polygon.n--;polygon.p[polygon.n] = polygon.p[0];return polygon;}void ChangeClockwise(TPolygon &polygon){TPoint tmp;int i;for(i = 0;i <= (polygon.n - 1) / 2;i++){tmp = polygon.p[i];polygon.p[i] = polygon.p[polygon.n - 1 - i];polygon.p[polygon.n - 1 - i] = tmp;}}double polygonArea(TPolygon p){    //已知多边形各顶点的坐标，求其面积    double area;    int i, n;    n = p.n;    area = 0;    for(i = 1;i <= n;i++){        area += (p.p[i - 1].x * p.p[i % n].y - p.p[i % n].x * p.p[i - 1].y);    }      return area / 2;   }int main(){int i, j;TPolygon polygon, new_polygon;while(scanf("%d", &polygon.n) && polygon.n){for(i = 0;i <= polygon.n - 1;i++){scanf("%lf%lf", &polygon.p[i].x, &polygon.p[i].y);}/*若是逆时针转化为顺时针*/if(polygonArea(polygon) > 0) ChangeClockwise(polygon);polygon.p[polygon.n] = polygon.p[0];new_polygon = polygon;for(i = 0;i <= polygon.n - 1;i++){new_polygon = Cut_polygon(polygon.p[i], polygon.p[i + 1], new_polygon);}if(new_polygon.n > 0) printf("1\n");else printf("0\n");}return 0;}

1.16.8 pku_2600_二分+圆的参数方程

#include <stdio.h>#include <math.h>const double eps = 1e-4;const double pi = acos(-1.0);struct TPoint {double x, y;}p[60], a[60];double angle[60];double multi(TPoint p1, TPoint p2, TPoint p0){    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);}TPoint fine_a2(TPoint a1, TPoint m, double angle1){TPoint a2;double r, angle2, angle3;r = sqrt((a1.x - m.x) * (a1.x - m.x) + (a1.y - m.y) * (a1.y - m.y));angle2 = acos((a1.x - m.x) / r);if(a1.y < m.y) {if(angle2 <= pi / 2) angle2 = -angle2;if(angle2 > pi / 2) angle2 = 3 * pi / 2 - (angle2 - pi / 2);}angle3 = angle2 - angle1;a2.x = m.x + r * cos(angle3);a2.y = m.y + r * sin(angle3);if(multi(m, a2, a1) < 0) return a2;    angle3 = angle2 + angle1;a2.x = m.x + r * cos(angle3);a2.y = m.y + r * sin(angle3);if(multi(m, a2, a1) < 0) return a2;}int main(){int n, i, j;while(scanf("%d", &n) != EOF){for(i = 0;i < n;i++){scanf("%lf%lf", &p[i].x, &p[i].y);}for(i = 0;i < n;i++){scanf("%lf", &angle[i]);angle[i] = angle[i] * pi / 180;}a[0].x = 0;a[0].y = 0;while(1){for(i = 1;i <= n;i++){a[i] = fine_a2(a[i - 1], p[i - 1], angle[i - 1]);}if(fabs(a[n].x - a[0].x) <= eps    && fabs(a[n].y - a[0].y) <= eps) break;else {a[0].x = (a[0].x + a[n].x) / 2;a[0].y = (a[0].y + a[n].y) / 2;}}for(i = 0;i < n;i++){printf("%.0lf %.0lf\n", a[i].x, a[i].y);}}return 0;}1.16.9 pku_1151_矩形相交的面积/*大牛的思想 题目给出 n 个矩形，要求它们的面积并。具体做法是离散化。先把 2n 个 x 坐标排序去重，然后再把所有水平线段（要记录是矩形上边还是下边）按 y 坐标排序。最后对于每一小段区间 (x[i], x[i + 1]) 扫描所有的水平线段，求出这些水平线段在小区间内覆盖的面积。总的时间复杂度是 O(n^2)。利用线段树，可以优化到 O(nlogn)。*/#include <stdio.h>#include <math.h>#include <stdlib.h>#define up  1#define down -1typedef struct TSeg{double l, r;double y;int UpOrDown;}TSeg;TSeg seg[210];int segn;double x[210];int xn;int cmp1(const void *a, const void *b){if(*(double *)a < *(double *)b) return -1;else return 1;}int cmp2(const void *a, const void *b){TSeg *c = (TSeg *)a;TSeg *d = (TSeg *)b;if(c->y < d->y) return -1;else return 1;}void movex(int t, int &xn){int i;for(i = t;i <= xn - 1;i++){x[i] = x[i + 1];}xn--;}int main(){//freopen("in.in", "r", stdin);//freopen("out.out", "w", stdout);int n, i, j, cnt, test = 1;double x1, y1, x2, y2, ylow, area;while(scanf("%d", &n) != EOF && n){xn = 0;segn = 0;for(i = 0;i < n;i++){scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);x[xn++] = x1;x[xn++] = x2;seg[segn].l = x1;seg[segn].r = x2;seg[segn].y = y1;seg[segn++].UpOrDown = up;seg[segn].l = x1;seg[segn].r = x2;seg[segn].y = y2;seg[segn++].UpOrDown = down;}qsort(x, xn, sizeof(x[0]), cmp1);/*除掉重复的x*/for(i = 1;i < xn;){if(x[i] == x[i - 1]) movex(i, xn);else i++;}qsort(seg, segn, sizeof(seg[0]), cmp2);area = 0.0;for(i = 0;i < xn - 1;i++){cnt = 0;for(j = 0;j < segn;j++){if(seg[j].l <= x[i] && seg[j].r >= x[i + 1]){if(cnt == 0) ylow = seg[j].y;if(seg[j].UpOrDown == down) cnt++;else cnt--;if(cnt == 0) area += (x[i + 1] - x[i]) * (seg[j].y - ylow);}}}printf("Test case #%d\n", test++);        printf("Total explored area: %.2lf\n", area); }return 0;}

1.16.10 pku_1118_共线最多的点的个数

/*2617120 chenhaifeng 1118 Accepted 512K 1890MS C++ 977B 2007-09-04 18:43:26 直接O(n^3)超时，用一个标记数组，标记i,j所做直线已经查找过，可以跳过大牛的思想朴素做法是 O(n3) 的，超时。我的做法是枚举每个点，然后求其它点和它连线的斜率，再排序。这样就得到经过该点的直线最多能经过几个点。求个最大值就行了。复杂度是 O(n2logn) 的。把排序换成 hash，可以优化到 O(n2)。 2617134 chenhaifeng 1118 Accepted 276K 312MS G++ 1394B 2007-09-04 18:49:08 */#include <stdio.h>#include <math.h>bool f[705][705];int a[705];int main(){int n, i, j, s, num, maxn;int x[705], y[705];int t, m;while(scanf("%d", &n) != EOF && n){for(i = 0;i <= n - 1;i++){scanf("%d%d", &x[i], &y[i]);}maxn = -1;for(i = 0;i <= n - 1;i++){for(j = i;j <= n - 1;j++){f[i][j] = false;}}for(i = 0;i <= n - 1;i++){for(j = i + 1;j <= n - 1;j++){if(f[i][j] == true) continue;if(n - j < maxn) break;num = 2;t = 2;a[0] = i;a[1] = j;f[i][j] = true; for(s = j + 1;s <= n - 1;s++){if(f[i][s] == true || f[j][s] == true) continue;if((y[i] - y[s]) * (x[j] - x[s]) == (x[i] - x[s]) * (y[j] - y[s])){ num++; a[t] = s; for(m = 0;m <= t - 1;m++){f[m][s] = true;}t++;} }if(num > maxn) maxn = num;}}printf("%d\n", maxn);}return 0;}

1.16.11 pku2826_线段围成的区域可储水量

/*两条线不相交，左边或右边的口被遮住，交点是某条线的那个纵坐标较高的那点某条线段水平放置*/#include <stdio.h>#include <math.h>#define eps 1e-8struct TPoint{double x, y;};struct TLine{    double a, b, c;};int same(TPoint p1, TPoint p2){if(fabs(p1.x - p2.x) > eps) return 0;if(fabs(p1.y - p2.y) > eps) return 0;return 1;}double min(double x, double y){    if(x < y) return x;    else return y; }double max(double x, double y){    if(x > y) return x;    else return y; }double multi(TPoint p1, TPoint p2, TPoint p0){     return (p1.x - p0.x) * (p2.y - p0.y)      - (p2.x - p0.x) * (p1.y - p0.y);}bool isIntersected(TPoint s1, TPoint e1, TPoint s2, TPoint e2){    if(    (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&    (max(s2.x, e2.x) >= min(s1.x, e1.x)) &&    (max(s1.y, e1.y) >= min(s2.y, e2.y)) &&    (max(s2.y, e2.y) >= min(s1.y, e1.y)) &&    (multi(s2, e1, s1) * multi(e1, e2, s1) >= 0) &&    (multi(s1, e2, s2) * multi(e2, e1, s2) >= 0)    )  return true;        return false;    }TLine lineFromSegment(TPoint p1, TPoint p2){    TLine tmp;    tmp.a = p2.y - p1.y;    tmp.b = p1.x - p2.x;    tmp.c = p2.x * p1.y - p1.x * p2.y;    return tmp;}TPoint LineInter(TLine l1, TLine l2){    TPoint tmp;     double a1 = l1.a;    double b1 = l1.b;    double c1 = l1.c;    double a2 = l2.a;    double b2 = l2.b;    double c2 = l2.c;    if(fabs(b1) < eps){        tmp.x = -c1 / a1;          tmp.y = (-c2 - a2 * tmp.x) / b2;    }           else{        tmp.x = (c1 * b2 - b1 * c2) / (b1 * a2 - b2 * a1);        tmp.y = (-c1 - a1 * tmp.x) / b1;    }return tmp;}double triangleArea(TPoint p1, TPoint p2, TPoint p3){TPoint p4, p5;p4.x = p2.x - p1.x;p4.y = p2.y - p1.y;p5.x = p3.x - p1.x;p5.y = p3.y - p1.y;return fabs(p5.x * p4.y - p5.y * p4.x) / 2;}double find_x(double y, TLine line){return (-line.c - line.b * y) / line.a;}double find_y(double x, TLine line){if(fabs(line.b) < eps){return -1e250;}else {return (-line.c - line.a  * x) / line.b;}}int main(){//freopen("in.in", "r", stdin);//freopen("out.out", "w", stdout);int test;double miny, y;TLine l1, l2;TPoint p1, p2, p3, p4, inter;TPoint tp1, tp2;scanf("%d", &test);while(test--){scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y, &p4.x, &p4.y);if(same(p1, p2) || same(p3, p4)    || !isIntersected(p1, p2, p3, p4)   || fabs(p1.y - p2.y) < eps //平行与x轴    || fabs(p3.y - p4.y) < eps  ){printf("0.00\n");continue;}l1 = lineFromSegment(p1, p2);l2 = lineFromSegment(p3, p4);inter = LineInter(l1, l2);if(p1.y > p2.y) tp1 = p1;else tp1 = p2;if(p3.y > p4.y) tp2 = p3;else tp2 = p4;if(tp1.y < tp2.y){if(tp1.x >= min(p4.x, p3.x) && tp1.x <= max(p4.x, p3.x)){y = find_y(tp1.x, l2);if(y >= tp1.y){printf("0.00\n");continue;}}miny = tp1.y;}else{if(tp2.x >= min(p1.x, p2.x) && tp2.x <= max(p1.x, p2.x)){y = find_y(tp2.x, l1);if(y >= tp2.y){printf("0.00\n");continue;}}miny = tp2.y;}if(fabs(miny - inter.y) < eps){printf("0.00\n");continue;}tp1.x = find_x(miny, l1);tp2.x = find_x(miny, l2);tp1.y = tp2.y = miny;printf("%.2lf\n", triangleArea(tp1, tp2, inter));} return 0;}

1.16.12 Pick公式

//A = b / 2 + i -1  其中 b 与 i 分別表示在边界上及內部的格子点之个數//http://www.hwp.idv.tw/bbs1/htm/%A6V%B6q%B7L%BFn%A4%C0/%A6V%B6q%B7L%BFn%A4%C0.htm// http://acm.pku.edu.cn/JudgeOnline/problem?id=2954#include <stdio.h>#include <stdlib.h>typedef struct TPoint{int x;int y;}TPoint;typedef struct TLine{int a, b, c;}TLine;int triangleArea(TPoint p1, TPoint p2, TPoint p3){    //已知三角形三个顶点的坐标，求三角形的面积     int k = p1.x * p2.y + p2.x * p3.y + p3.x * p1.y      - p2.x * p1.y - p3.x * p2.y - p1.x * p3.y;      if(k < 0) return -k;    else return k;}TLine lineFromSegment(TPoint p1, TPoint p2){    //线段所在直线,返回直线方程的三个系统     TLine tmp;    tmp.a = p2.y - p1.y;    tmp.b = p1.x - p2.x;    tmp.c = p2.x * p1.y - p1.x * p2.y;    return tmp;}void swap(int &a, int &b){int t;t = a;a = b;b = t;}int Count(TPoint p1, TPoint p2){int i, sum = 0, y;TLine l1 =  lineFromSegment(p1, p2);if(l1.b == 0) return abs(p2.y - p1.y) + 1;if(p1.x > p2.x) swap(p1.x, p2.x); //这里没有交换WA两次 for(i = p1.x;i <= p2.x;i++){y = -l1.c - l1.a * i;if(y % l1.b == 0) sum++;}return sum;}int main(){ //freopen("in.in", "r", stdin);//freopen("OUT.out", "w", stdout);TPoint p1, p2, p3;while(scanf("%d%d%d%d%d%d", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y) != EOF){if(p1.x == 0 && p1.y == 0 && p2.x == 0 && p2.y == 0 && p3.x == 0 && p3.y == 0) break;int  A = triangleArea(p1, p2, p3);//A为面积的两倍 int b = 0;int i;b = Count(p1, p2) + Count(p1, p3) + Count(p3, p2) - 3;//3个顶点多各多加了一次 //i = A / 2- b / 2 + 1;i = (A - b) / 2 + 1;printf("%d\n", i);} return 0;}

1.16.13 N点中三个点组成三角形面积最大

//Rotating Calipers algorithm#include <stdio.h>#include <stdlib.h>#include <math.h>#define MaxNode 50005int stack[MaxNode];int top;double max;typedef struct TPoint{    int x;    int y;}TPoint;TPoint point[MaxNode];void swap(TPoint point[], int i, int j){    TPoint tmp;    tmp = point[i];    point[i] = point[j];    point[j] = tmp;}double multi(TPoint p1, TPoint p2, TPoint p0){    return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);}double distance(TPoint p1, TPoint p2){return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y);}int cmp(const void *a, const void *b){    TPoint *c = (TPoint *)a;    TPoint *d = (TPoint *)b;    double k = multi(*c, *d, point[0]);    if(k< 0) return 1;    else if(k == 0 && distance(*c, point[0]) >= distance(*d, point[0]))          return 1;    else return -1;   }void grahamScan(int n){     //Graham扫描求凸包    int i, u;         //将最左下的点调整到p[0]的位置    u = 0;    for(i = 1;i <= n - 1;i++){        if((point[i].y < point[u].y) ||      (point[i].y == point[u].y && point[i].x  < point[u].x))u = i;          }     swap(point, 0, u);        //将平p[1]到p[n - 1]按按极角排序，可采用快速排序    qsort(point + 1, n - 1, sizeof(point[0]), cmp);        for(i = 0;i <= 2;i++) stack[i] = i;    top = 2;    for(i = 3;i <= n - 1;i++){        while(multi(point[i], point[stack[top]], point[stack[top - 1]]) >= 0){top--;if(top == 0) break;        }        top++;        stack[top] = i;    }}int main(){        double triangleArea(int i, int j, int k);    void PloygonTriangle();    int i, n;    while(scanf("%d", &n) && n != -1){        for(i = 0;i < n;i++)scanf("%d%d", &point[i].x, &point[i].y);        if(n <= 2){printf("0.00\n");            continue;              }        if(n == 3){            printf("%.2lf\n", triangleArea(0, 1, 2));            continue;         }        grahamScan(n);        PloygonTriangle();        printf("%.2lf\n", max);        }    return 0;}void PloygonTriangle(){    double triangleArea(int i, int j, int k);    int i, j , k;    double area, area1;    max = -1;for(i = 0;i <= top - 2;i++){ k = -1;for(j = i + 1; j <= top - 1;j++){ if(k <= j) k= j + 1;area = triangleArea(stack[i], stack[j], stack[k]);if(area > max) max = area;while(k + 1 <= top){area1= triangleArea(stack[i], stack[j], stack[k + 1]);if(area1 < area) break;if(area1 > max) max = area1;area = area1;k++;}}}}double triangleArea(int i, int j, int k){    //已知三角形三个顶点的坐标，求三角形的面积     double l = fabs(point[i].x * point[j].y + point[j].x * point[k].y + point[k].x * point[i].y - point[j].x * point[i].y - point[k].x * point[j].y - point[i].x * point[k].y) / 2;      return l;}

1.16.14 直线关于圆的反射

/*fzu_10351.直线和圆的交点 2.点关于线的对称点3.点到线的距离 4.直线方程 */#include <iostream>#include <cmath>using namespace std;#define INF 999999999const double eps = 1e-6;int up;typedef struct TPoint{    double x;    double y;}TPoint;typedef struct TCircle{    TPoint center;    double r;}TCircle;typedef struct TLine{    //直线标准式中的系数     double a, b, c;}TLine;void SloveLine(TLine &line, TPoint start, TPoint dir){    //根据直线上一点和直线的方向求直线的方程     if(dir.x == 0){        line.a = 1;        line.b = 0;        line.c = start.x;    }    else {        double k = dir.y / dir.x;        line.a = k;        line.b = -1;        line.c = start.y - k * start.x;            }    }TLine lineFromSegment(TPoint p1, TPoint p2){    //线段所在直线,返回直线方程的三个系统     TLine tmp;    tmp.a = p2.y - p1.y;    tmp.b = p1.x - p2.x;    tmp.c = p2.x * p1.y - p1.x * p2.y;    return tmp;}TPoint symmetricalPointofLine(TPoint p, TLine L){    //p点关于直线L的对称点     TPoint p2;    double d;    d = L.a * L.a + L.b * L.b;    p2.x = (L.b * L.b * p.x - L.a * L.a * p.x -             2 * L.a * L.b * p.y - 2 * L.a * L.c) / d;    p2.y = (L.a * L.a * p.y - L.b * L.b * p.y -             2 * L.a * L.b * p.x - 2 * L.b * L.c) / d;    return p2;}double distanc(TPoint p1, TPoint p2){    //计算平面上两个点之间的距离   return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));    }bool samedir(TPoint dir, TPoint start, TPoint point){    //判断方向     TPoint tmp;    tmp.x = point.x - start.x;    tmp.y = point.y - start.y;    if(tmp.x != 0 && dir.x != 0){        if(tmp.x / dir.x > 0) return true;        else return false;    }    else if(tmp.y != 0 && dir.y != 0){        if(tmp.y / dir.y > 0) return true;        else return false;     }    return true; }bool Intersected(TPoint &point, TLine line, const TCircle circle[],  TPoint start, TPoint dir, int which){    //如果圆与直线有(有效交点)交点就存放在变量point中 double a = line.a, b = line.b, c = line.c;double x0 = circle[which].center.x, y0 = circle[which].center.y;double r =  circle[which].r;//有交点，求交点 double x2front = b * b + a * a;double x1front = -2 * x0 * b * b + 2 * a * b * y0 + 2 * a * c;double front = x0 * x0 * b * b + y0 * y0 * b * b + c * c + 2 * c * y0 * b - b * b * r * r;double d = x1front * x1front - 4 * x2front * front;TPoint p1, p2;bool k1, k2;    if(fabs(d) < eps){        //x2front不可能等于零        point.x = -x1front / x2front / 2;       point.y = (-c - a * point.x) / b;       //判断方向       if(samedir(dir, start, point)) return true;       else return false;    }    else if(d < 0) return false;    else {        p1.x = (-x1front + sqrt(d)) / 2 / x2front;        p1.y = (-c - a * p1.x) / b;        p2.x = (-x1front - sqrt(d)) / 2 / x2front;        p2.y = (-c - a * p2.x) / b;        k1 = samedir(dir, start, p1);        k2 = samedir(dir, start, p2);        if(k1 == false && k2 == false) return false;        if(k1 == true && k2 == true){            double dis1 = distanc(p1, start);            double dis2 = distanc(p2, start);            if(dis1 < dis2) point = p1;            else point = p2;            return true;        }        else if(k1 == true) point = p1;        else point = p2;        return true;    }}void Reflect(int &num, TCircle circle[], TPoint start, TPoint dir, int n){    //反复反射     int i;    TLine line;    TPoint interpoint, newstart;    int u;    SloveLine(line, start, dir);    int tag = 0;    double mindis = INF;    for(i = 1;i <= n;i++){        if(i != up && Intersected(interpoint, line, circle, start, dir, i)){            double dis = distanc(start, interpoint);            if(dis < mindis){                tag = 1;                u = i;                  mindis = dis;                newstart = interpoint;            }                    }    }    if(tag == 0){        cout << "inf" << endl;        return ;    }    else {        if(num == 10){            cout << "..." << endl;           return ;        }        cout << u << " ";         num++;        //新的方向         TLine line1;        TPoint p;        line1 = lineFromSegment(newstart, circle[u].center);        if(fabs(line1.a * start.x + line1.b * start.y +line1.c) <= eps){            dir.x = -dir.x;            dir.y = -dir.y;        }        else {            p = symmetricalPointofLine(start, line1);//start的对称点             dir.x = p.x - newstart.x;            dir.y = p.y - newstart.y;         }                start = newstart;             up = u;        Reflect(num, circle, start, dir, n);         } }int main(){    //freopen("fzu_1035.in", "r", stdin);    //freopen("fzu_1035.out", "w", stdout);    int n, i, j, num, test = 1;    TCircle circle[30];    TPoint start, dir;    while(cin >> n && n){        for(i = 1;i <= n;i++){            cin >> circle[i].center.x >> circle[i].center.y >> circle[i].r;        }        cin >> start.x >> start.y >> dir.x >> dir.y;                cout << "Scene " << test++ << endl;        num = 0;        up = -1;        Reflect(num, circle, start, dir, n);        cout << endl;    }     return 0;}

1.16.15 pku2002_3432_N个点最多组成多少个正方形(hao)

#include <stdio.h> #include <stdlib.h> #include <math.h> #define eps 1e-6#define pi acos(-1.0)#define PRIME 9991struct point{     int x, y; }p[2201]; int n;struct HASH{    int cnt;    int next;}hash[50000];int hashl;int Hash(int n){    int i = n % PRIME;    while(hash[i].next != -1){        if(hash[hash[i].next].cnt == n) return 1;        else if(hash[hash[i].next].cnt > n) break;        i = hash[i].next;    }    hash[hashl].cnt = n;    hash[hashl].next = hash[i].next;    hash[i].next = hashl;    hashl++;     return 0;}int Hash2(int n){    int i = n % PRIME;    while(hash[i].next != -1){        if(hash[hash[i].next].cnt == n) return 1;        else if(hash[hash[i].next].cnt > n) return 0;        i = hash[i].next;    }    return 0;}int check(double ax, double ay, int &x, int &y){    int a0 = (int)ax;    int b0 = (int)ay;    int tag1 = 0, tag2 = 0;       if(fabs(a0 - ax) < eps){        tag1 = 1;        x = a0;    }    else if(fabs(a0 + 1 - ax) < eps){        tag1 = 1;        x = a0 + 1;    }    if(fabs(b0 - ay) < eps){        tag2 = 1;        y = b0;    }    else if(fabs(b0 + 1 - ay) < eps){        y = b0 + 1;        tag2 = 1;    }    if(tag1 == 1 && tag2 == 1) return 1;    else return 0;}int squares(point p1, point p2, point &p3, point &p4){    double a = (double)p2.x - p1.x;    double b = (double)p2.y - p1.y;    double midx = ((double)p1.x + p2.x) / 2;    double midy = ((double)p1.y + p2.y) / 2;    double tmp = a * a + b * b;    double x1 = sqrt(b * b) / 2;    double y1;    if(fabs(b) < eps) y1 = sqrt(a * a + b * b) / 2;    else y1 = -a * x1 / b;    x1 += midx;    y1 += midy;    if(check(x1, y1, p3.x, p3.y) == 0) return 0;        x1 = 2 * midx - x1;    y1 = 2 * midy - y1;     if(check(x1, y1, p4.x, p4.y) == 0) return 0;     return 1;}int main() {     int i, j, cnt;     while(scanf("%d", &n) != EOF && n)    {for(i = 0;i < PRIME;i++) hash[i].next = -1;hashl = PRIME;int x1, y1, x2, y2;for (i = 0; i < n; i++){scanf("%d%d", &p[i].x, &p[i].y); Hash((p[i].x + 100000) * 100000 + p[i].y + 100000);}   cnt = 0; for (i = 0; i < n; i++){ for (j = i + 1; j < n; j++) { point a, b; if(squares(p[i], p[j], a, b) == 0) continue; if(Hash2((a.x + 100000) * 100000 + a.y + 100000) == 0) continue;if(Hash2((b.x + 100000) * 100000 + b.y + 100000) == 0) continue;cnt++; } }printf("%d\n", cnt / 2); }    return 0; }

1.16.16 pku1981_单位圆覆盖最多点(poj1981)CircleandPoints

/*平面上N个点，用一个半径R的圆去覆盖，最多能覆盖多少个点？比较经典的题目。对每个点以R为半径画圆，对N个圆两两求交。这一步O(N^2)。问题转化为求被覆盖次数最多的弧。对每一个圆，求其上的每段弧重叠次数。假如A圆与B圆相交。A上[PI/3, PI/2]的区间被B覆盖(PI为圆周率)。那么对于A圆，我们在PI/3处做一个+1标记，在PI/2处做一个-1标记。对于[PI*5/3, PI*7/3]这样横跨0点的区间只要在0点处拆成两段即可。将一个圆上的所有标记排序，从头开始扫描。初始ans = 0，碰到+1标记给ans++，碰到-1标记ans--。扫描过程中ans的最大值就是圆上被覆盖最多的弧。求所有圆的ans的最大值就是答案。总复杂度O(N^2 * logN)*/#include <stdio.h>#include <math.h>#define eps 1e-6struct point {double x, y;};double dis(point p1, point p2){point p3;p3.x = p2.x - p1.x;p3.y = p2.y - p1.y;return p3.x * p3.x + p3.y * p3.y;}point find_centre(point p1, point p2){point p3, mid, centre;double b, c, ang;p3.x = p2.x - p1.x;p3.y = p2.y - p1.y;mid.x = (p1.x + p2.x) / 2;mid.y = (p1.y + p2.y) / 2;b = dis(p1, mid);c = sqrt(1 - b);if(fabs(p3.y) < eps)//垂线的斜角90度{centre.x = mid.x;centre.y = mid.y + c;}else {ang = atan(-p3.x / p3.y);centre.x = mid.x + c * cos(ang);centre.y = mid.y + c * sin(ang);}return centre;}int main(){    int n, ans, tmpans, i, j, k;    point p[305], centre;    double tmp;    while(scanf("%d", &n) && n)    {for(i = 0;i < n;i++)scanf("%lf%lf", &p[i].x, &p[i].y);ans = 1;for(i = 0;i < n;i++)for(j = i + 1;j < n;j++){if(dis(p[i], p[j]) > 4) continue;tmpans = 0;centre = find_centre(p[i], p[j]);for(k = 0;k < n;k++){//if(tmpans + n - k <= ans) break;tmp = dis(centre, p[k]);//if(tmp < 1.0 || fabs(tmp - 1.0) < eps) tmpans++;if(tmp <= 1.000001) tmpans++;}if(ans < tmpans) ans = tmpans;}printf("%d\n", ans);    }    return 0;}

1.16.17 pku3668_GameofLine_N个点最多确定多少互不平行的直线(poj3668)

#include <math.h>#include <stdio.h>#include <stdlib.h>#define eps 1e-6#define pi acos(-1)struct point {double x, y;};double FindSlewRate(point p1, point p2){point p;p.x = p2.x - p1.x;p.y = p2.y - p1.y;if(fabs(p.x) < eps) return pi / 2;double tmp = atan(p.y / p.x);if(tmp < 0) return pi + tmp;return tmp;}int cmp(const void *a, const void *b){double *c = (double *)a;double *d = (double *)b;if(*c < *d) return -1;return 1;}int main(){int n, rt;point p[205];double rate[40005];while(scanf("%d", &n) != EOF){for(int i = 0;i < n;i++)scanf("%lf%lf", &p[i].x ,&p[i].y);rt = 0;for(int i = 0;i < n;i++)for(int j = i + 1;j < n;j++)rate[rt++] = FindSlewRate(p[i], p[j]);qsort(rate, rt, sizeof(rate[0]), cmp);int ans = 1;for(int i = 1;i < rt;i++)if(rate[i] > rate[i - 1]) ans++;//注意这里写fabs(rate[i] - rate[i - 1]) > eps Wrong Answer printf("%d\n", ans);} return 0;}

1.16.18 求凸多边形直径

#include<stdio.h> #include<math.h> #define eps 1e-6 #define MaX 6000 /*-----------多边形结构------------*/ struct POLYGON{    int n;                                  //多边形顶点数    double x[MaX],y[MaX];        //顶点坐标 }poly; int zd[100000][2],znum;      //跖对的集合和跖对的数量 /*------------辅助函数-----------*/ double dist(int a,int b,int c) {    double vx1,vx2,vy1,vy2;    vx1=poly.x[b]-poly.x[a]; vy1=poly.y[b]-poly.y[a];    vx2=poly.x[c]-poly.x[a]; vy2=poly.y[c]-poly.y[a];    return fabs(vx1*vy2 - vy1*vx2); } /*-------------求凸多边形直径的函数-------------*/ double DIAMETER() {    znum=0;    int i,j,k=1;    double m,tmp;    while(dist(poly.n-1,0,k+1) > dist(poly.n-1,0,k)+eps)       k++;    i=0; j=k;    while(i<=k && j<poly.n)    {       zd[znum][0]=i; zd[znum++][1]=j;       while(dist(i,i+1,j+1)>dist(i,i+1,j)-eps && j<poly.n-1)       {          zd[znum][0]=i; zd[znum++][1]=j;          j++;       }       i++;    }    m=-1;    for(i=0;i<znum;i++)    {       tmp =(poly.x[zd[i][0]]-poly.x[zd[i][1]]) * (poly.x[zd[i][0]]-poly.x[zd[i][1]]);       tmp+=(poly.y[zd[i][0]]-poly.y[zd[i][1]]) * (poly.y[zd[i][0]]-poly.y[zd[i][1]]);       if(m<tmp) m=tmp;    }    return sqrt(m); } /*----------主函数----------*/ int main() {    int i;    while(scanf("%d",&poly.n)==1)    {       for(i=0;i<poly.n;i++)          scanf("%lf %lf",&poly.x[i],&poly.y[i]);       printf("%.3lf\n",DIAMETER());    }    return 0; }1.16.19 矩形面积并，周长并见附录一5.11,5.121.16.20 pku2069 最小球覆盖 见附录一5.13,5.14//最小闭合球#include<stdio.h>#include<math.h>#include<memory>#include<stdlib.h>using namespace std;const double eps = 1e-10;struct point_type { double x, y, z; };int npoint, nouter ;point_type point [1000], outer[4], res;double radius, tmp ;inline double dist(point_type p1 , point_type p2){double dx=p1.x-p2.x, dy=p1.y-p2.y,dz=p1.z-p2.z ;return ( dx*dx + dy*dy + dz*dz ) ;}inline double dot( point_type p1 , point_type p2 ){return p1.x*p2.x + p1.y*p2.y + p1.z*p2.z;}void ball(){point_type q[3]; double m[3][3],sol[3],L[3],det; int i,j;res.x=res.y=res.z=-1000;radius=0;    switch ( nouter ) {case 1 : res=outer[0]; break;case 2 : res.x=(outer[0].x+outer[1].x)/2; res.y=(outer[0].y+outer[1].y)/2; res.z=(outer[0].z+outer[1].z)/2; radius=dist(res,outer[0]); break;case 3 : for ( i=0; i<2; ++i ) {q[i].x=outer[i+1].x-outer[0].x;q[i].y=outer[i+1].y-outer[0].y;q[i].z=outer[i+1].z-outer[0].z;} for ( i=0; i<2; ++i ) for ( j=0; j<2; ++j )m[i][j]=dot(q[i],q[j])*2 ; for ( i=0; i<2; ++i ) sol[i]=dot(q[i],q[i]); if (fabs(det=m[0][0]*m[1][1]-m[0][1]*m[1][0]) < eps ) return ;  L[0]=(sol[0]*m[1][1]-sol[1]*m[0][1])/det; L[1]=(sol[1]*m[0][0]-sol[0]*m[1][0])/det; res.x=outer[0].x+q[0].x*L[0]+q[1].x*L[1]; res.y=outer[0].y+q[0].y*L[0]+q[1].y*L[1]; res.z=outer[0].z+q[0].z*L[0]+q[1].z*L[1]; radius=dist(res,outer[0]); break;case 4 : for ( i=0; i<3; ++i ){q[i].x=outer[i+1].x-outer[0].x;q[i].y=outer[i+1].y-outer[0].y;q[i].z=outer[i+1].z-outer[0].z;sol[i]=dot(q[i],q[i]);} for ( i=0; i<3; ++i)  for ( j=0; j<3; ++j) m[i][j]=dot(q[i],q[j])*2; det= m[0][0]*m[1][1]*m[2][2] + m[0][1]*m[1][2]*m[2][0] +m[0][2]*m[2][1]*m[1][0] - m[0][2]*m[1][1]*m[2][0] -m[0][1]*m[1][0]*m[2][2] - m[0][0]*m[1][2]*m[2][1]; if ( fabs( det )<eps ) return; for ( j=0; j<3; ++j ){for ( i=0; i<3; ++i ) m[i][j]=sol[i]; L[j]=( m[0][0]*m[1][1]*m[2][2] + m[0][1]*m[1][2]*m[2][0]  + m[0][2]*m[2][1]*m[1][0] - m[0][2]*m[1][1]*m[2][0]  - m[0][1]*m[1][0]*m[2][2] - m[0][0]*m[1][2]*m[2][1]  ) / det;for( i=0; i<3; ++i ) m[i][j]=dot(q[i],q[j])*2;} res=outer[0]; for ( i=0; i<3; ++i ) {res.x+=q[i].x*L[i];    res.y+=q[i].y*L[i];res.z+=q[i].z*L[i];} radius=dist(res,outer[0]);}}void minball(int n){  ball();  if ( nouter <4 )       for ( int i=0; i<n; ++i )           if( dist(res,point[i])-radius>eps)    {outer[nouter]=point[i];                ++nouter;minball(i);--nouter;if(i>0){  point_type Tt = point[i] ;  memmove(&point[1], &point[0] , sizeof ( point_type )*i );  point[0]=Tt;}}}int main(){ int i; while(scanf("%d",&npoint)!=EOF,npoint){ for(i=0;i<npoint;i++)       scanf("%lf%lf%lf",&point[i].x,&point[i].y,&point[i].z); nouter=0;  minball(npoint); printf("%.8lf\n",sqrt(radius)+eps);} return 0;}

1.16.21 最大空凸包、最大空矩形

见附录一5.7, 6.15

1.16.22 求圆和多边形的交

/*圆和简单多边形*/#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>#include <iostream>#include <ctime>using namespace std;#define M 30#define eps 1e-7const double PI = acos(-1.0);class pnt_type{public:double x,y;};class state_type{public:double angle;double CoverArea;};pnt_type pnt[M];pnt_type center;int n;double R;bool read_data(){n = 3;int i;if (cin >> pnt[1].x >> pnt[1].y){for (i=2;i<=n;i++) cin >> pnt[i].x >> pnt[i].y;cin >> center.x >> center.y >> R;return true;}return false;}inline double Area2(pnt_type &a,pnt_type &b,pnt_type &c){return (b.x - a.x) * (c.y - a.y) - (c.x - a.x) * (b.y - a.y);}inline double dot(pnt_type &a,pnt_type &b,pnt_type &c){return (b.x - a.x) * (c.x - a.x) + (b.y - a.y) * (c.y - a.y);}inline double dist(pnt_type &a,pnt_type &b){return sqrt((b.x - a.x) * (b.x - a.x) + (b.y - a.y) * (b.y - a.y));}void init(){int i;double temp,sum;for (i=2;i<n;i++){temp = Area2(pnt[1],pnt[i],pnt[i + 1]);sum += temp;}if (sum < 0) reverse(pnt + 1,pnt + n + 1);pnt[n + 1] = pnt[1];}inline bool inCircle(pnt_type &s){return dist(center,s) <= R;}bool SameSide(pnt_type a,pnt_type b){if (dist(a,center) > dist(b,center)) swap(a,b);return dot(a,b,center) < eps;}double ShadomOnCircle(pnt_type a,pnt_type b){double flag = Area2(center,a,b),res = 0;if (fabs(flag) < eps) return 0;bool ina = inCircle(a),inb = inCircle(b);if (ina && inb){res = fabs(Area2(center,a,b)) / 2;}else if (!ina && !inb){if (SameSide(a,b)){double theta = acos(dot(center,a,b) / dist(center,a) / dist(center,b));res = R * R * theta / 2;}else{double height = fabs(Area2(center,a,b)) / dist(a,b);double theta = acos(dot(center,a,b) / dist(center,a) / dist(center,b));if (height >= R){res = R * R * theta / 2;}else{double _theta = 2 * acos(height / R);res = R * R * (theta - _theta) / 2 + R * R * sin(_theta) / 2;}}}else{if (!ina && inb) swap(a,b);double height = fabs(Area2(center,a,b)) / dist(a,b);double temp = dot(a,center,b);double theta = acos(dot(center,a,b) / dist(center,a) / dist(center,b)),theta1,theta2;if (fabs(temp) < eps){double _theta = acos(height / R);res += R * height / 2 * sin(_theta);res += R * R / 2 * (theta - _theta);}else{theta1 = asin(height / R); theta2 = asin(height / dist(a,center));if (temp > 0){res += dist(center,a) * R / 2 * sin(PI - theta1 - theta2);res += R * R / 2 * (theta + theta1 + theta2 - PI);}else{res += dist(center,a) * R / 2 * sin(theta2 - theta1);res += R * R / 2 * (theta - theta2 + theta1);}}}if (flag < 0) return -res; else return res;}double Cover(){int i;double res = 0;for (i=1;i<=n;i++) res += ShadomOnCircle(pnt[i],pnt[i + 1]);return res;}int main(){double ans;while (read_data()){init();ans = Cover();printf("%.2lf\n",ans);}return 0;}

半平面交

//Nlgn#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;#define maxn 20005#define eps 1e-10struct point{double x,y;};struct line{point s,e;double k;};line L[maxn];point S[maxn];int N,Q[maxn];double cross(point a,point b,point c) // c在直线ab右边返回<0{return (c.y-a.y)*(b.x-a.x)-(b.y-a.y)*(c.x-a.x);}bool operator < (line a,line b) // 直线se右边为可行域{if( fabs(a.k-b.k)<eps )return cross(b.s,b.e,a.s)<0;return a.k<b.k;}point intersection(point u1,point u2,point v1,point v2){point ret=u1;double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));ret.x+=(u2.x-u1.x)*t;ret.y+=(u2.y-u1.y)*t;return ret;}double HalfInSec(){int i,j,k,l;sort(L,L+N);  // 极角[-pi,pi]排序for(i=1,j=0;i<N;i++)  // 去掉相同极角直线if( fabs(L[i].k-L[j].k)>eps ) L[++j] = L[i];N = j+1;k = 0,l = 1;Q[0] = 0,Q[1] = 1;S[1] = intersection(L[0].s,L[0].e,L[1].s,L[1].e);for(i=2;i<N;i++){while( k<l && cross(L[i].s,L[i].e,S[l])>eps ) l--;while( k<l && cross(L[i].s,L[i].e,S[k+1])>eps ) k++;Q[++l] = i;S[l] = intersection(L[Q[l-1]].s,L[Q[l-1]].e,L[i].s,L[i].e);}while( k<l && cross(L[Q[k]].s,L[Q[k]].e,S[l])>eps )l--;while( k<l && cross(L[Q[l]].s,L[Q[l]].e,S[k+1])>eps )k++;S[k] = intersection(L[Q[l]].s,L[Q[l]].e,L[Q[k]].s,L[Q[k]].e);S[++l] = S[k];double s = 0;for(i=k;i<l;i++)s += S[i].y*S[i+1].x-S[i+1].y*S[i].x;return fabs(s/2);}int main(){int i,j,k,l;scanf("%d",&N);for(i=0;i<N;i++)scanf("%lf%lf%lf%lf",&L[i].e.x,&L[i].e.y,&L[i].s.x,&L[i].s.y);L[N].s.x = 0,L[N].s.y = 0;L[N+1].s.x = 10000,L[N+1].s.y = 0;L[N+2].s.x = 10000,L[N+2].s.y = 10000;L[N+3].s.x = 0,L[N+3].s.y = 10000;L[N].e = L[N+3].s;L[N+1].e = L[N].s;L[N+2].e = L[N+1].s;L[N+3].e = L[N+2].s;N += 4;for(i=0;i<N;i++)L[i].k = atan2(L[i].s.y-L[i].e.y,L[i].s.x-L[i].e.x);printf("%.1lf\n",HalfInSec());}

//N^2#include <iostream>#include <math.h>#include <string>using namespace std;#define maxn 2005int N;struct point{double x, y;}P[maxn];double xmul(point a,point b,point c){return (c.y-a.y)*(b.x-a.x)-(b.y-a.y)*(c.x-a.x);}point intersection(point u1,point u2,point v1,point v2){point ret=u1;double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));ret.x+=(u2.x-u1.x)*t;ret.y+=(u2.y-u1.y)*t;return ret;}int HalfInteSec(double move){int i,j;point tt[maxn],tp[maxn],ns,ne;int len,tlen;len = N;memcpy(tp,P,sizeof(P));for(i=0;i<N;i++){point s,e;double dx,dy;dx = P[i].y-P[i+1].y;dy = P[i+1].x-P[i].x;dx = P[i].y-P[i+1].y;dy = P[i+1].x-P[i].x;s.x = P[i+1].x+dx*(move/sqrt(dx*dx+dy*dy));s.y = P[i+1].y+dy*(move/sqrt(dx*dx+dy*dy));e.x = P[i].x+dx*(move/sqrt(dx*dx+dy*dy));e.y = P[i].y+dy*(move/sqrt(dx*dx+dy*dy));tlen = 0;for(j=0;j<len;j++){ns = tp[j];    ne = tp[j+1];if( xmul(s,e,ns)<=0 )tt[tlen++] = ns;if( xmul(s,e,ns)*xmul(s,e,ne)<0 )tt[tlen++] = intersection(s,e,ns,ne);}tt[tlen] = tt[0];memcpy(tp,tt,sizeof(tt));len = tlen;}return len;}int main(){int i,j,k,l;while( scanf("%d",&N),N ){for(i=0;i<N;i++)scanf("%lf%lf",&P[i].x,&P[i].y);P[N] = P[0];double min = 0,max = 10000,mid;while( max-min>0.0000001 ){mid = (min+max)/2;l = HalfInteSec(mid);if( l==0 )max = mid;elsemin = mid;}printf("%.6lf\n",min);}}