You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8452    Accepted Submission(s):
4125

Problem Description

Many geometry（几何）problems were designed in the
ACM/ICPC. And now, I also prepare a geometry problem for this final exam.
According to the experience of many ACMers, geometry problems are always much
trouble, but this problem is very easy, after all we are now attending an exam,
not a contest :)
Give you N (1<=N<=100) segments（线段）, please output the
number of all intersections（交点）. You should count repeatedly if M (M>2)
segments intersect at the same point.

Note:
You can assume that two
segments would not intersect at more than one point.

Input

Input contains multiple test cases. Each test case
contains a integer N (1=N<=100) in a line first, and then N lines follow.
Each line describes one segment with four float values x1, y1, x2, y2 which are
coordinates of the segment’s ending.
A test case starting with 0 terminates
the input and this test case is not to be processed.

Output

For each case, print the number of intersections, and
one line one case.

Sample Input

2

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.00

3

0.00 0.00 1.00 1.00

0.00 1.00 1.00 0.000

0.00 0.00 1.00 0.00

0

Sample Output

1

3

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <stack>
 5 #include <queue>
 6 #include <map>
 7 #include <set>
 8 #include <vector>
 9 #include <math.h>
10 #include <algorithm>
11 using namespace std;
12 const double pi = acos(-1.0);
13 const int INF = 0x3f3f3f3f;
14
15 struct Line
16 {
17     double x1,y1,x2,y2;
18 }node[105];
19
20 bool solve(Line a, Line b)//叉积判断两线段是否相交
21 {
22     if (((a.x1-b.x1)*(a.y2-b.y1)-(a.x2-b.x1)*(a.y1-b.y1))*((a.x1-b.x2)*(a.y2-b.y2)-(a.x2-b.x2)*(a.y1-b.y2))>0)
23     return false;
24     if (((b.x1-a.x1)*(b.y2-a.y1)-(b.x2-a.x1)*(b.y1-a.y1))*((b.x1-a.x2)*(b.y2-a.y2)-(b.x2-a.x2)*(b.y1-a.y2))>0)
25     return false;
26     return true;
27 }
28
29 int main ()
30 {
31     int n;
32     while (scanf ("%d",&n),n)
33     {
34         for (int i=0; i<n; i++)
35         scanf ("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2);
36         int cnt = 0;
37         for (int i=0; i<n; i++)
38         {
39             for (int j=i+1; j<n; j++)
40             {
41                 if (solve(node[i], node[j]))
42                 cnt++;
43             }
44         }
45         printf ("%d\n",cnt);
46     }
47     return 0;
48 }