【POJ】2318 TOYS ——计算几何+二分

 1 #include <cstdio>
 2 #include <cstring>
 3 
 4 const int LEN = 5050;
 5 
 6 struct Point //点的结构体
 7 {
 8     int x;
 9     int y;
10 };
11 
12 struct Line //线段的结构体
13 {
14     Point a;
15     Point b;
16 }line[LEN];
17 
18 int ans[LEN];
19 
20 int judge(Line tline, Point p3)  //运用叉积的性质判断点在线段的左边还是右边
21 {
22     Point t1, t2;
23     t1.x = p3.x - tline.b.x;
24     t1.y = p3.y - tline.b.y;
25     t2.x = tline.a.x - tline.b.x;
26     t2.y = tline.a.y - tline.b.y;
27     return t1.x * t2.y - t1.y * t2.x;
28 }
29 
30 int divide(int l, int r, Point toy) //二分查找
31 {
32     int mid = (l + r) / 2;
33     if (mid == l)
34         return l;
35     if (judge(line[mid], toy) < 0)
36         return divide(l, mid, toy);
37     else
38         return divide(mid, r, toy);
39 }
40 
41 int main()
42 {
43     int n, m, x1, y1, x2, y2;
44     //freopen("in.txt", "r", stdin);
45     while(scanf("%d", &n) != EOF && n){
46         scanf("%d %d %d %d %d", &m, &x1, &y1, &x2, &y2);
47         memset(ans, 0, sizeof(ans));
48         for(int i = 1; i <= n; i++){
49             int up, low;
50             scanf("%d %d", &up, &low);
51             line[i].a.x = up;
52             line[i].a.y = y1;
53             line[i].b.x = low;
54             line[i].b.y = y2;
55         }
56         n++;
57         line[0].a.x = line[0].b.x = x1;
58         line[0].b.y = line[n].b.y = y2;
59         line[0].a.y = line[n].a.y = y1;
60         line[n].a.x = line[n].b.x = x2;
61         for(int i = 0; i < m; i++){
62             Point toy;
63             scanf("%d %d", &toy.x, &toy.y);
64             ans[divide(0, n, toy)]++;
65         }
66         for(int i = 0; i < n; i++){
67             printf("%d: %d\n", i, ans[i]);
68         }
69         printf("\n");
70     }
71     return 0;
72 }