poj1410 Intersection

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题目链接：http://poj.org/problem?id=1410
题意：给你一条线段的两个端点坐标，再给你一个矩形的两个顶点坐标，问你这条线段是否和矩形相交，或者这条线段在矩形里面，如果是输出T否则输出F
解析：用线段相交的判断方法判断线段和矩形的四条边是否相交，然后用来类似跨立实验的东西判断线段是否在矩形里面

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <iostream>#include <vector>#include <queue>#include <set>using namespace std;struct point{    double x,y;    point() {}    point(double _x,double _y)    {        x = _x;        y = _y;    }};struct line{    point a,b;    line() {}    line(point _a,point _b)    {        a = _a;        b = _b;    }};double x_mul(point p0,point p1,point p2){    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}bool judge(line l1,line l2){    bool flag1 =  min(l1.a.x,l1.b.x)<=max(l2.a.x,l2.b.x)&&                min(l2.a.x,l2.b.x)<=max(l1.a.x,l1.b.x)&&                min(l1.a.y,l1.b.y)<=max(l2.a.y,l2.b.y)&&                min(l2.a.y,l2.b.y)<=max(l1.a.y,l1.b.y);    bool flag2 = (x_mul(l1.a,l2.a,l2.b)*x_mul(l1.b,l2.a,l2.b)<=0)&&                (x_mul(l2.a,l1.a,l1.b)*x_mul(l2.b,l1.a,l1.b)<=0);    return flag1&&flag2;}int main(){    int n;    scanf("%d",&n);    while(n--)    {        line t;        point a,b;        scanf("%lf %lf %lf %lf",&t.a.x,&t.a.y,&t.b.x,&t.b.y);        scanf("%lf %lf %lf %lf",&a.x,&a.y,&b.x,&b.y);        if(a.x>b.x)            swap(a.x,b.x);        if(a.y<b.y)            swap(a.y,b.y);        if(a.x<min(t.a.x,t.b.x)&&b.x>max(t.a.x,t.b.x)&&           a.y>max(t.a.y,t.b.y)&&b.y<min(t.a.y,t.b.y))            puts("T");        else        {            line t1 = line(point(a.x,b.y),point(b.x,b.y));            line t2 = line(point(a.x,a.y),point(b.x,a.y));            line t3 = line(point(a.x,a.y),point(a.x,b.y));            line t4 = line(point(b.x,a.y),point(b.x,b.y));            if(judge(t,t1)||judge(t,t2)||judge(t,t3)||judge(t,t4))                puts("T");            else                puts("F");        }    }    return 0;}

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