HDU 1086You can Solve a Geometry Problem too(判断两条选段是否有交点）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1086

判断两条线段是否有交点，我用的是跨立实验法：

两条线段分别是A1到B1，A2到B2，很显然，如果这两条线段有交点，那么可以肯定的是：

A1-B1,A2-B1这两个向量分别在B2-B1的两边，判断是不是在两边可以用向量的叉积来判断，这里就不说了，同理B1-A1,B2-A1在A2-A1的两边，当同时满足这两个条件时，说明这两条线段是有交点的。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 105;
const double eps = 1e-6;
struct point
{
    double x,y;
    point(double x = 0,double y = 0):x(x),y(y) {}
    inline friend point operator + (point p1,point p2)
    {
        return point(p1.x+p2.x,p1.y+p2.y);
    }
    inline friend point operator - (point p1,point p2)
    {
        return point(p1.x-p2.x,p1.y-p2.y);
    }
}A[maxn],B[maxn];

inline double dot(point p1,point p2)
{
    return p1.x*p2.y - p2.x*p1.y;
}
inline double dis(point p1,point p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
}
int judge(point p1,point p2,point p3,point p4)
{
    double temp = dot(p3-p1,p2-p1) * dot(p4-p1,p2-p1);
    if(temp < 0 || fabs(temp) < eps) return 1;
    return 0;
}

int main()
{
    //freopen("in","r",stdin);
    int n;
    while(scanf("%d",&n),n)
    {
        for(int i = 0;i < n;++i)
        scanf("%lf%lf%lf%lf",&A[i].x,&A[i].y,&B[i].x,&B[i].y);
        int ans = 0;
        for(int i = 0;i < n;++i)
        for(int j = i+1;j < n;++j)
        {
             if(judge(A[i],B[i],A[j],B[j]) && judge(A[j],B[j],A[i],B[i])) ans++;
        }
        printf("%d
",ans);
    }
    return 0;
}