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。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 const int maxn = 105;
 8 const double eps = 1e-6;
 9 struct point
10 {
11     double x,y;
12     point(double x = 0,double y = 0):x(x),y(y) {}
13     inline friend point operator + (point p1,point p2)
14     {
15         return point(p1.x+p2.x,p1.y+p2.y);
16     }
17     inline friend point operator - (point p1,point p2)
18     {
19         return point(p1.x-p2.x,p1.y-p2.y);
20     }
21 }A[maxn],B[maxn];
22
23 inline double dot(point p1,point p2)
24 {
25     return p1.x*p2.y - p2.x*p1.y;
26 }
27 inline double dis(point p1,point p2)
28 {
29     return sqrt((p1.x-p2.x)*(p1.x-p2.x) + (p1.y-p2.y)*(p1.y-p2.y));
30 }
31 int judge(point p1,point p2,point p3,point p4)
32 {
33     double temp = dot(p3-p1,p2-p1) * dot(p4-p1,p2-p1);
34     if(temp < 0 || fabs(temp) < eps) return 1;
35     return 0;
36 }
37
38 int main()
39 {
40     //freopen("in","r",stdin);
41     int n;
42     while(scanf("%d",&n),n)
43     {
44         for(int i = 0;i < n;++i)
45         scanf("%lf%lf%lf%lf",&A[i].x,&A[i].y,&B[i].x,&B[i].y);
46         int ans = 0;
47         for(int i = 0;i < n;++i)
48         for(int j = i+1;j < n;++j)
49         {
50              if(judge(A[i],B[i],A[j],B[j]) && judge(A[j],B[j],A[i],B[i])) ans++;
51         }
52         printf("%d\n",ans);
53     }
54     return 0;
55 }