1 node A, B;
 2 struct line
 3 {
 4     node a, b;
 5 }L[100];
 6 int cross_product(node A, node B, node C)
 7 {
 8     return ((A.x - B.x) * (A.y - C.y) - (A.y - B.y) * (A.x - B.x));
 9 }
10 if (cross_product(A, L[i].a, L[i].b) * cross_product(B, L[i].a, L[i].b) < -EPS) // 判断是否相交
11 {
12     ++ cnt;
13 }

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <cmath>
 6 #include <stack>
 7 #include <map>
 8 #include <queue>
 9 #include <set>
10 #include <climits>
11
12 using namespace std;
13 const int MAXN = 35;
14 const int MY_MAX = INT_MAX;
15 const int EPS = 1e-8;
16 int N, n;
17
18 struct node
19 {
20     double x, y;
21 };
22
23 struct line
24 {
25     node a, b;
26 }l[MAXN];
27
28 int cross_product(node A, node B, node C)
29 {
30     return ((B.x - A.x) * (C.y - A.y) - (B.y - A.y) * (C.x - A.x));
31 }
32
33 int solve(node K, node A)
34 {
35     int cnt = 1;
36     for(int i = 0; i < n; ++ i)
37     {
38         if (cross_product(A, l[i].a, l[i].b) * cross_product(K, l[i].a, l[i].b) < -EPS)
39             ++ cnt;
40     }
41     return cnt;
42 }
43
44 int main()
45 {
46     scanf("%d", &N);
47     while(N --)
48     {
49         int cnt = MY_MAX;
50         scanf("%d", &n);
51         for(int i = 0; i < n; ++ i)
52             scanf("%lf%lf%lf%lf", &l[i].a.x, &l[i].a.y, &l[i].b.x, &l[i].b.y);
53         node k;
54         scanf("%lf%lf", &k.x, &k.y);
55         if (n == 0)
56             printf("1\n");
57         else
58         {
59             for(int i = 0; i < n; ++ i)
60             {
61                 cnt = min(cnt, solve(k, l[i].a));
62                 cnt = min(cnt, solve(k, l[i].b));
63             }
64             printf("%d\n", cnt);
65         }
66     }
67     return 0;
68 }