计算几何基础

1.多边形面积计算

double S(Point p[],int n){    double ans = 0;    p[n] = p[0];    for(int i=1;i<n;i++)       ans += cross(p[0],p[i],p[i+1]);    if(ans < 0) ans = -ans;    return ans / 2.0;}

2.求凸包

bool cmp(Point A,Point B){    double k = cross(MinA,A,B);    if(k<0) return 0;    if(k>0) return 1;    return dist(MinA,A)<dist(MinA,B);}void Graham(Point p[],int n){    for(int i=1;i<n;i++)       if(p[i].y<p[0].y || (p[i].y == p[0].y && p[i].x < p[0].x))            swap(p[i],p[0]);    MinA = p[0];    p[n] = p[0];    sort(p+1,p+n,cmp);    stack[0] = p[0];    stack[1] = p[1];    top = 2;    for(int i=2;i<n;i++)    {        while(top >= 2 && cross(stack[top-2],stack[top-1],p[i])<=0) top--;        stack[top++] = p[i];    }}

3.任意多边形求重心

Point Gravity(Point p[],int n){    Point O,t;    O.x = O.y = 0;    t.x = t.y = 0;    p[n] = p[0];    double A = 0;    for(int i=0; i<n; i++)        A += cross(O,p[i],p[i+1]);    A /= 2.0;    for(int i=0; i<n; i++)    {        t.x += (p[i].x + p[i+1].x) * cross(O,p[i],p[i+1]);        t.y += (p[i].y + p[i+1].y) * cross(O,p[i],p[i+1]);    }    t.x /= 6*A;    t.y /= 6*A;    return t;}

4.求线段交点的坐标

bool Segment_crossing(Segment u,Segment v)   /** 判断线段是否相交 */{         return((max(u.a.x,u.b.x)>=min(v.a.x,v.b.x))&&           (max(v.a.x,v.b.x)>=min(u.a.x,u.b.x))&&           (max(u.a.y,u.b.y)>=min(v.a.y,v.b.y))&&           (max(v.a.y,v.b.y)>=min(u.a.y,u.b.y))&&           (cross(v.a,u.b,u.a)*cross(u.b,v.b,u.a)>=0)&&           (cross(u.a,v.b,v.a)*cross(v.b,u.b,v.a)>=0));}/**求直线交点的坐标，如果没有交点返回NULL，否则返回交点p的地址*/Point* CrossPoint(Segment u,Segment v){    Point p;    if(Segment_crossing(u,v))    {        p.x=(cross(v.b,u.b,u.a)*v.a.x-cross(v.a,u.b,u.a)*v.b.x)/(cross(v.b,u.b,u.a)-cross(v.a,u.b,u.a));        p.y=(cross(v.b,u.b,u.a)*v.a.y-cross(v.a,u.b,u.a)*v.b.y)/(cross(v.b,u.b,u.a)-cross(v.a,u.b,u.a));        return &p;    }    return NULL;}

5.三角形外接圆的半径与圆心

Point Circle_Point(Point A,Point B,Point C){    double a=dist(B,C);    double b=dist(A,C);    double c=dist(A,B);    double p=(a+b+c)/2.0;    double S=sqrt(p*(p-a)*(p-b)*(p-c));    R=(a*b*c)/(4*S);    //三角形外接圆的半径为R    double t1=(A.x*A.x+A.y*A.y-B.x*B.x-B.y*B.y)/2;    double t2=(A.x*A.x+A.y*A.y-C.x*C.x-C.y*C.y)/2;    Point center;    center.x=(t1*(A.y-C.y)-t2*(A.y-B.y))/((A.x-B.x)*(A.y-C.y)-(A.x-C.x)*(A.y-B.y));    center.y=(t1*(A.x-C.x)-t2*(A.x-B.x))/((A.y-B.y)*(A.x-C.x)-(A.y-C.y)*(A.x-B.x));    return center;}

6.旋转卡壳求凸包的直径，也就是平面最远点对，p[]为凸包的点集

double rotating_calipers(Point p[],int n){    int k = 1;    double ans = 0;    p[n] = p[0];    for(int i=0;i<n;i++)    {        while(fabs(cross(p[i],p[i+1],p[k])) < fabs(cross(p[i],p[i+1],p[k+1])))             k = (k+1) % n;        ans = max(ans, max(dist(p[i],p[k]),dist(p[i+1],p[k])));    }    return ans;}

7.求凸包的宽度

double rotating_calipers(Point p[],int n){    int k = 1;    double ans = 0x7FFFFFFF;    p[n] = p[0];    for(int i=0;i<n;i++)    {        while(fabs(cross(p[i],p[i+1],p[k])) < fabs(cross(p[i],p[i+1],p[k+1])))             k = (k+1) % n;        double tmp = fabs(cross(p[i],p[i+1],p[k]));        double d   = dist(p[i],p[i+1]);        ans = min(ans,tmp/d);    }    return ans;}

8. 求线段与圆的交点

#include <iostream>#include <stdio.h>#include <vector>#include <math.h>using namespace std;const double eps = 1e-8;struct Point{    double x, y;};struct Segment{    Point s, t;};struct Circle{    Point c;    double r;};vector<Point> LineToCircle(Segment line, Circle circle){    vector<Point> v;    v.clear();    double fd = sqrt((line.s.x - line.t.x) * (line.s.x - line.t.x)                   + (line.s.y - line.t.y) * (line.s.y - line.t.y));    Point d;    d.x = (line.t.x - line.s.x) / fd;    d.y = (line.t.y - line.s.y) / fd;    Point e;    e.x = circle.c.x - line.s.x;    e.y = circle.c.y - line.s.y;    double a = e.x * d.x + e.y * d.y;    double a2 = a * a;    double e2 = e.x * e.x + e.y * e.y;    double r2 = circle.r * circle.r;    v.push_back(line.s);    if(r2 - e2 + a2 > -eps)    {        double f = sqrt(r2 - e2 + a2);        double t = a - f;        if((t > -eps) && (t - fd) < eps)        {            Point tmp;            tmp.x = line.s.x + t * d.x;            tmp.y = line.s.y + t * d.y;            v.push_back(tmp);        }        t = a + f;        if((t > -eps) && (t - fd) < eps)        {            Point tmp;            tmp.x = line.s.x + t * d.x;            tmp.y = line.s.y + t * d.y;            v.push_back(tmp);        }    }    v.push_back(line.t);    return v;}int main(){    Segment line;    Circle circle;    line.s.x = -1.0;    line.s.y = -1.0;    line.t.x = 1.0;    line.t.y = 1.0;    circle.c.x = 0.0;    circle.c.y = 0.0;    circle.r = 2.0;    vector<Point> p = LineToCircle(line, circle);    for(int i = 0; i < p.size(); i++)        printf("%lf %lf\n", p[i].x, p[i].y);    return 0;}

现在有这样一个问题，平面上给定n个点，确定一对平行线，使得所有的点都在平行线之间，求这对平行线的最近距

离。这个问题实际上就是求点集的凸包宽度。