POJ2398 Toy Storage（点与凸多边形位置关系）

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <vector>
 6 #include <cmath>
 7 #include <utility>
 8 using namespace std;
 9 
10 typedef pair<int, int> pii;
11 
12 const int N = 1010;
13 
14 struct Point {
15     double x, y;
16     Point(double x = 0, double y = 0) :x(x), y(y) {}
17 };    //点的定义
18 
19 typedef Point Vector;    //向量的定义
20 
21 Vector operator - (Vector& A, Vector& B) { return Vector(A.x - B.x, A.y - B.y); }    //向量减法
22 
23 double Cross(Vector& A, Vector& B) { return A.x*B.y - A.y*B.x; }    //向量叉乘
24 
25 int cnt[N], X1, Y1, X2, Y2, n, m, ans[N];
26 pii P[N];
27 Vector ve[N];
28 
29 int calc(Point& p, int l, int r) {    //二分
30     if (l == r)return l;
31     int mid = (l + r) >> 1;
32     Vector tmp(p.x - P[mid].second, p.y - Y2);
33     if (Cross(tmp, ve[mid]) < 0)return calc(p, l, mid);
34     else return calc(p, mid + 1, r);
35 }
36 
37 int main() {
38     while (cin >> n&&n) {
39         memset(cnt, 0, sizeof(cnt));
40         memset(ans, 0, sizeof(ans));
41         scanf("%d%d%d%d%d", &m, &X1, &Y1, &X2, &Y2);
42         ve[n].x = 0, ve[n].y = Y1 - Y2;
43         P[n].first = X2, P[n].second = X2;
44         int dy = Y1 - Y2;
45         for (int i = 0; i < n; ++i)
46             scanf("%d%d", &P[i].first, &P[i].second);
47         sort(P, P + n + 1);    //排序
48         for (int i = 0; i <= n; ++i)
49             ve[i].x = P[i].first - P[i].second, ve[i].y = dy;
50         Point tmp;
51         for (int i = 0; i < m; ++i) {
52             scanf("%lf%lf", &tmp.x, &tmp.y);
53             ++cnt[calc(tmp, 0, n)];
54         }
55         for (int i = 0; i <= n; ++i)
56             ++ans[cnt[i]];
57         printf("Box\n");
58         for (int i = 1; i <= n; ++i)if (ans[i])
59             printf("%d: %d\n", i, ans[i]);
60     }
61 }