hdu 2826(好坑,线段相交问题)

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
const double eps = 1e-8;
struct Point
{
    double x,y;
};
double cross(Point a,Point b,Point c)
{
    return (a.x-c.x)*(b.y-c.y)-(a.y-c.y)*(b.x-c.x);
}
///规范相交
bool isCross(Point a,Point b,Point c,Point d)
{
    if(cross(c,b,a)*cross(b,d,a)<-eps) return false; ///这里要改成eps我上面的那组数据才能为0.5..不过是0也能AC。。so strange
    if(cross(a,d,c)*cross(d,b,c)<-eps) return false;
    return true;
}
///计算两条直线的交点
Point intersection(Point a,Point b,Point c,Point d)
{
    Point p = a;
    double t = ((a.x-c.x)*(c.y-d.y)-(a.y-c.y)*(c.x-d.x))/((a.x-b.x)*(c.y-d.y)-(a.y-b.y)*(c.x-d.x));
    p.x +=(b.x-a.x)*t;
    p.y +=(b.y-a.y)*t;
    return p;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        Point a,b,c,d;
        scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
        scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
        if(a.y==b.y||c.y==d.y||!isCross(a,b,c,d))  ///排除水平放置还有不相交的情况
        {
            printf("0.00\n");
            continue;
        }
        Point p = intersection(a,b,c,d); ///交点
        double y = min(max(a.y,b.y),max(c.y,d.y));
        if(y<=p.y)  ///上面的y不可能小于交点,不然接不到水
        {
            printf("0.00\n");
            continue;
        }
        ///我只要y上面的点
        Point t1,t2;
        if(a.y>b.y) t1 = a;
        else t1 = b;
        if(c.y>d.y) t2 = c;
        else t2 = d;
        ///两个向量极角大的x坐标必定小于极角小的，不然雨水没办法流进去
        if(cross(t1,t2,p)>0&&t1.x>t2.x||cross(t2,t1,p)>0&&t2.x>t1.x)
        {
            double k,B,x,x0;
            if(y==t1.y)
            {
                x = t1.x;
                if(t2.x==p.x) ///这里略坑
                {
                    x0 = p.x;
                }
                else
                {
                    k = (t2.y- p.y)/(t2.x - p.x);
                    B = t2.y-k*t2.x;
                    x0 = (y-B)/k;
                }
            }
            else
            {
                x = t2.x;
                if(t1.x==p.x)
                {
                    x0 = p.x;
                }
                else
                {
                    k = (t1.y- p.y)/(t1.x - p.x);
                    B = t1.y-k*t1.x;
                    x0 = (y-B)/k;
                }

            }
            double l = fabs(x-x0);
            double h = fabs(y-p.y);
            printf("%.2lf\n",l*h/2);
            continue;
        }
        printf("0.00\n");
    }
    return 0;
}