poj 1410 Intersection

1、文中给出的左上顶点和右下顶点不保证x1<x2,y1>y2;即需要自己判断

2、文中似乎没说，但必须这么认为：线段完全在矩形内部要返回T.

3.判断两个线段相交时，注意它们在一条直线上的情况

#include<stdio.h>
struct Point
{
int x,y;
}pointc,point1,point2,point[5];
struct Line
{
Point a,b;
}linex;
int 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 dmult(Point p1,Point p2,Point p0)
{
return (p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);
}
int main()
{
int i,j,_case;
scanf("%d",&_case);
while(_case--)
{
scanf("%d%d%d%d",&linex.a.x,&linex.a.y,&linex.b.x,&linex.b.y);
scanf("%d%d%d%d",&point1.x,&point1.y,&point2.x,&point2.y);
if(point1.x>point2.x)
{
pointc.x=point2.x;
point2.x=point1.x;
point1.x=pointc.x;
}
if(point1.y>point2.y)
{
pointc.y=point2.y;
point2.y=point1.y;
point1.y=pointc.y;
}
if(linex.a.x>=point1.x&&linex.a.x<=point2.x&&
linex.a.y>=point1.y&&linex.a.y<=point2.y)//至少有一个点在矩形中
{
printf("T\n");
continue;
}
if(linex.b.x>=point1.x&&linex.b.x<=point2.x&&
linex.b.y>=point2.y&&linex.b.y<=point1.y)//至少有一个点在矩形中
{
printf("T\n");
continue;
}
point[0]=point1;
point[1].x=point1.x;point[1].y=point2.y;
point[2]=point2;
point[3].x=point2.x;point[3].y=point1.y;
point[4]=point1;
for(i=0;i<4;i++)
{
int xm1=xmult(linex.a,point[i],point[i+1])*xmult(linex.b,point[i],point[i+1]);
int xm2=xmult(point[i],linex.a,linex.b)*xmult(point[i+1],linex.a,linex.b);
if(xm1<=0&&xm2<=0)
{
//printf("%d %d %d %d\n",point[i].x,point[i].y,point[i+1].x,point[i+1].y);
if(xm1==0&&xm2==0)
{
if(dmult(linex.a,linex.b,point[i])<=0)break;
if(dmult(linex.a,linex.b,point[i+1])<=0)break;
if(dmult(point[i],point[i+1],linex.a)<=0)break;
if(dmult(point[i],point[i+1],linex.b)<=0)break;
continue;
}
break;
}
}
if(i==4)printf("F\n");
else printf("T\n");
}
return 0;
}

官方测试数据：