poj 1410 Intersection
**必须注意几点
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：
