HDU-Rectangle and Circle

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这个是一个模板题，谢谢岩儿提供的模板。

疑问：直线到点的距离模板 和 线段的到点距离的模板有什么区别？应该怎么写？

Trick：我自己使用的一个for循环中每一个边必须是矩形的边而不应该是对角线，所以pp数组的顺序是固定的。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<algorithm>

using namespace std;

const double eps = 1e-10;

typedef struct

{

double x, y;

}point;

double dist(point p1,point p2)

{

return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.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);

}

//点到直线的距离

double disptoline(point p,point l1,point l2)

{

return fabs(xmult(p,l1,l2))/dist(l1,l2);

}

//相交返回1，不想交返回0

int intersect_seg_circle(point c,double r, point l1,point l2)

{

double t1=dist(c,l1)-r,t2=dist(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;

}

point pp[4];

point cc;

int main()

{

int Case;

double X, Y, R, X1, Y1, X2, Y2;

cin>>Case;

while( Case-- )

{

cin>>X>>Y>>R>>X1>>Y1>>X2>>Y2;

pp[0].x = X1;

pp[0].y = Y1;

pp[1].x = X1;

pp[1].y = Y2;

pp[2].x = X2;

pp[2].y = Y2;

pp[3].x = X2;

pp[3].y = Y1;

cc.x = X;

cc.y = Y;

bool flag = false;

for(int i = 0; i < 4; i++)

{

if( intersect_seg_circle(cc, R, pp[i % 4], pp[(i+1) % 4]) ) { flag = true; break; }

}

if( flag ) cout<<"YES"<<endl;

else cout<<"NO"<<endl;

}

return 0;

}