POJ 1039 Pipe

题意：光线从管子射入，求在管子中能传播的最远的距离。若能穿过管子则输出"Through all the pipe."

题解：刘汝佳书上的一道题。注意看题··。题中的一条光线同时经过几个折点也可以···

#include<cstdio>#include<cmath>#include<cstdlib>#include<algorithm>#define eps 1e-8#define zero(x) (((x)>0?(x):-(x))<eps)using std::max;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;}int n;Point top[100], bot[100];bool cal ( Point pl, Point pr, double &res ){    if ( xmult(pr,top[1],pl) < -eps || xmult(pr,bot[1],pl) > eps )        return false;    int i;    bool f1 = true, f2 = true;    for ( i = 2; i <= n; i++ )    {        if ( xmult(pr,top[i],pl) < -eps) f1 = false;        if ( xmult(pr,bot[i],pl) > eps ) f2 = false;        if ( !f1 || !f2 ) break;    }    if ( f1 && f2 ) return true;    Line l; Point tmp;    l.a.x = top[1].x; l.b.x = top[n].x;    l.a.y = (pl.y-pr.y)/(pl.x-pr.x)*(l.a.x-pr.x)+pr.y;    l.b.y = (pl.y-pr.y)/(pl.x-pr.x)*(l.b.x-pl.x)+pl.y;    if ( ! f1 )    {        tmp = intersection(l.a,l.b, top[i-1], top[i]);        res = max(res,tmp.x);    }    if ( ! f2 )    {        tmp = intersection(l.a,l.b, bot[i-1], bot[i]);        res = max(res,tmp.x);    }    return false;}int main(){    while ( scanf("%d",&n) && n )    {        for ( int i = 1; i <= n; i++ )        {            scanf("%lf %lf", &top[i].x, &top[i].y);            bot[i].x = top[i].x;            bot[i].y = top[i].y - 1;        }        bool pass = false;        double res = -999999999999.99;        for ( int i = 1; i < n; i++ )        {            for ( int j = i + 1; j <= n; j++ )            {                if ( cal(top[i],bot[j],res) ) { pass = true; break; }                if ( cal(bot[i],top[j],res) ) { pass = true; break; }            }        }        if ( pass == true )            printf("Through all the pipe.\n");        else printf("%.2lf\n",res);    }    return 0;}