poj 1410

题目不难，就是线段相交判断。叉积和点积的应用，但是却需要细心，看清题意

#include <iostream>
#include <stdio.h>
using namespace std;
int cross(int x1,int y1,int x2,int y2)
{
	return x1*y2-y1*x2;
}
int dot(int x1,int y1,int x2,int y2)
{
	return x1*x2+y1*y2;
}
int ons(int x1,int y1,int x2,int y2,int x3,int y3)
{
	if(cross(x3-x1,y3-y1,x3-x2,y3-y2)==0&&dot(x3-x1,y3-y1,x3-x2,y3-y2)<=0) return 1;
	else return 0;
}
int judge(int x1,int y1,int x2,int y2,int x3,int y3,int x4,int y4)
{
	int c1=cross(x2-x1,y2-y1,x3-x1,y3-y1),c2=cross(x2-x1,y2-y1,x4-x1,y4-y1),
		c3=cross(x4-x3,y4-y3,x1-x3,y1-y3),c4=cross(x4-x3,y4-y3,x2-x3,y2-y3);
	if(c1*c2<0&&c3*c4<=0)
		return 1;
	else if(c1*c2==0&&(ons(x1,y1,x2,y2,x3,y3)||ons(x1,y1,x2,y2,x4,y4)))
		return 1;
	return 0;
}
int main()
{
	int sx,sy,ex,ey,xleft,ytop,xright,ybottom;
	int n;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d%d%d%d%d%d%d",&sx,&sy,&ex,&ey,&xleft,&ytop,&xright,&ybottom);
		int minx=min(xleft,xright),maxx=max(xleft,xright);
		int miny=min(ytop,ybottom),maxy=max(ytop,ybottom);
		int flag=0;
		if((ex>=minx&&ex<=maxx&&sx>=minx&&sx<=maxx&&sy>=miny&&sy<=maxy&&ey>=miny&&ey<=maxy)||judge(sx,sy,ex,ey,xleft,ytop,xright,ytop)||judge(sx,sy,ex,ey,xleft,ytop,xleft,ybottom)
			||judge(sx,sy,ex,ey,xleft,ybottom,xright,ybottom)||judge(sx,sy,ex,ey,xright,ybottom,xright,ytop))
			flag=1;
		if(flag) printf("T\n");
		else printf("F\n");
	}
	return 0;
}