UVA 11529 - Strange Tax Calculation（计数问题)

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题目链接：11529 - Strange Tax Calculation

题意：平面上n个建筑物，3个建筑物可以组成一个三角形，计算平均每个三角形内有多少个点

思路：问题等价于，求凹四边形的占所有四边形的比例，用O(n^2)的算法，跟

HDU 3629 Convex

这题是一个道理

代码：

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;const double eps = 1e-9;const double pi = acos(-1.0);const int N = 1225;int t, n, tn;struct Point {double x, y;void read() {scanf("%lf%lf", &x, &y); }} p[N];double r[2 * N];double C(int n, int m) {if (m > n) return 0;double ans = 1;for (int i = 0; i < m; i++)ans = ans * (n - i) / (i + 1); return ans;}double cal(Point a, Point b) {return atan2(b.y - a.y, b.x - a.x);}double solve(int num) {tn = 0;double ans = 0;for (int i = 0; i < n; i++) {if (i == num) continue;r[tn++] = cal(p[num], p[i]); } sort(r, r + tn);int j = 1;for (int i = 0; i < tn; i++)r[i + tn] = r[i] + 2 * pi;  for (int i = 0; i < tn; i++) {  while (fabs(r[j] - r[i]) - pi < -eps) j++;  ans += C(j - i - 1, 2);}return C(tn, 3) - ans;}int main() {int cas = 0;while (scanf("%d", &n) && n) {double ans = 0;for (int i = 0; i < n; i++)p[i].read();for (int i = 0; i < n; i++) {ans += solve(i);  }  printf("City %d: %.2lf\n", ++cas, ans / C(n, 3)); }return 0;}

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