POJ 2318

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 struct Point 
 6 {
 7     int x, y;
 8 };
 9 
10 struct Line 
11 {
12    Point a, b;
13 } line[5005];
14 
15 int ans[5005];                //答案数组 
16 
17 int Multi(Point p1, Point p2, Point p0)         //p2相对p1左转为正 ，叉积 
18 {
19     return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
20 }
21 
22 void BSearch(Point a, int n) 
23 {
24     int l, r, mid;
25 
26     l = 0; r = n-1;
27     while (l < r) 
28     {
29         mid = (l + r) >> 1;        //  /2
30         if (Multi(a, line[mid].a, line[mid].b) > 0)         //.a为上边的节点 ,若a节点在盒子中 
31             l = mid + 1;                                    //左区间推进 
32         else 
33             r = mid;                                        //右区间推进 
34     }
35     if (Multi(a, line[l].a, line[l].b) < 0)         //判断边界 
36         ans[l]++;    
37     else 
38         ans[l+1]++;
39 }
40 
41 int main() 
42 {
43     int n, m, x1, y1, x2, y2;
44     int  t1, t2;
45     Point a;
46     while (scanf ("%d", &n) && n)                 //处理 0 结尾 数据 的 好技巧 
47     {
48         scanf ("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
49         for (int i = 0; i < n; i++) 
50         {
51             scanf ("%d%d", &t1, &t2);
52             line[i].a.x = t1;            //.a为上边的节点 
53             line[i].a.y = y1;
54             line[i].b.x = t2;
55             line[i].b.y = y2;
56         }
57         memset(ans, 0, sizeof (ans));
58         for (int i = 0; i < m; i++) 
59         {
60             scanf ("%d%d", &a.x, &a.y);
61             BSearch(a, n);
62         }
63         for (int i = 0; i <= n; i++)
64             printf ("%d: %d\n", i, ans[i]);
65         printf("\n");
66     }
67     return 0;
68 }