poj 1410 (没做出来，记得闲着没事看看这道题)

听说这道题是个大大的坑题

结果wa了十多发，，，，还是没找到原因

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
double eps=1e-8;
int cmp(double x)
{
    if(fabs(x)<eps)
        return 0;
    if(x<0)return -1;
    return 1;
}
 
struct Point
{
    double x,y;
    Point (){}
    Point (double _x,double _y)
    {
        x=_x,y=_y;
    }
    Point operator -(const Point &b)const
    {
    return Point (x-b.x,y-b.y);
    }
    double operator *(const Point &b)const
    {
        return x*b.x+y*b.y;
    }
    double operator ^(const Point &b)const
    {
        return x*b.y-y*b.x;
    }
};
 
struct Line
{
    Point s,e;
    Line (){}
    Line (Point _s,Point _e)
    {
        s=_s;
        e=_e;
    }
};
 
double xmult(Point p0,Point p1,Point p2)
{
    return cmp((p2-p1)^(p0-p1));
}
bool seg_seg(Line l1,Line l2)//相交return true
{
    return xmult(l1.s,l2.s,l2.e)*xmult(l1.e,l2.s,l2.e)<=0&& xmult(l2.s,l1.s,l1.e)*xmult(l2.e,l1.s,l1.e)<=0;
}
Line line[5];
 
int main ()
{
    int n;
    double x1,x2,x3,x4,y1,y2,y3,y4;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
        if(x3>x4)swap(x3,x4);
        if(y3<y4)swap(y3,y4);
        line[0]=Line(Point(x1,y1),Point(x2,y2));
        if(x1>x3&&x1<x4&&x2<x4&&x2>x3&&y1>y4&&y1<y3&&y2>y4&&y2<y3)
        {
                printf("T\n");
                continue;
        }
        else
        {
            line[1]=Line(Point(x3,y4),Point(x3,y3));
            line[2]=Line(Point(x3,y4),Point(x4,y4));
            line[3]=Line(Point(x4,y3),Point(x4,y4));
            line[4]=Line(Point(x4,y3),Point(x3,y3));
            for(int i=1;i<=4;i++)
            {
                if(seg_seg(line[0],line[i]))
                {
                    printf("T\n");//相交了
                    break;
                }
                if(i==4)
                    printf("F\n");//没相交
            }
        }
 
    }
 
    return 0;
}

已经发现的 坑点：

1：相交的意思是：与这个矩形整体相交，即分为：（1）与四条边中的几条相交，（2）或者在矩形内部不与四条边相交；

2：给出的矩形的四条边不一定是左上边的和右下边的，需要进行比较。（太坑）