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 = ;
 const double eps = 1e-;
 struct point
 {
     double x,y;
     point(double x = ,double y = ):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 <  || fabs(temp) < eps) return ;
     return ;
 }
 
 int main()
 {
     //freopen("in","r",stdin);
     int n;
     while(scanf("%d",&n),n)
     {
         for(int i = ;i < n;++i)
         scanf("%lf%lf%lf%lf",&A[i].x,&A[i].y,&B[i].x,&B[i].y);
         int ans = ;
         for(int i = ;i < n;++i)
         for(int j = i+;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\n",ans);
     }
     return ;
 }