poj1410 Intersection 线段与矩形相交

//poj1410
//题目意思：判断线段是否与矩形相交（注：线段在矩形内也是相交）
//解题思路：图形学中的线段裁剪算法，有Cohen-Sutherland,梁友栋-Barsky等，
//这里用容易理解的方法：先判断是否有一个点在矩形内，若都没有再去判断4条边与线段是否相交
//0ms AC 代码如下：

#include<iostream>
#include<cstdio>
#include<math.h>

#define eps 1e-8
struct Point{double x,y;};
bool zero(double x)
{
	return x>0? x<eps: x>-eps;
}
double xmult(Point p1,Point p2,Point p0)
{
	return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
int dot_online_in(Point p,Point l1,Point l2)//点在线段上
{
	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 same_side(Point p1,Point p2,Point p3,Point p4)
{
	return xmult(p1,p3,p4)*xmult(p2,p3,p4)>eps;
}
int intersect_in(Point p1,Point p2,Point p3,Point p4)//线段相交
{
	if(!zero(xmult(p1,p2,p3))&&!zero(xmult(p1,p2,p4)))
		return !same_side(p1,p2,p3,p4)&&!same_side(p3,p4,p1,p2);
	return dot_online_in(p1,p3,p4)||dot_online_in(p2,p3,p4)
		||dot_online_in(p3,p1,p2)||dot_online_in(p4,p1,p2);
}
bool judge(Point p0,Point p1,Point p2)
{
	if((p1.x-p0.x)*(p2.x-p0.x)<=0.0 &&(p1.y-p0.y)*(p2.y-p0.y)<=0.0)return true;
	return false;
}
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		Point start,end,p[4];
		scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&start.x,&start.y,&end.x,&end.y,&p[0].x,&p[0].y,&p[2].x,&p[2].y);
		p[1].x=p[0].x;p[1].y=p[2].y;
		p[3].x=p[2].x;p[3].y=p[0].y;
		int i;
		if(judge(start,p[0],p[2])||judge(end,p[0],p[2]))
		{
			printf("T\n");
			continue;
		}
		else
		{
			for(i=0;i<4;i++)
			{
				if(intersect_in(start,end,p[i],p[(i+1)%4]))
				{
					printf("T\n");
					break;
				}
			}
			if(i==4)printf("F\n");
		}
	}
	return 0;
}