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;
}

UVA 11529 - Strange Tax Calculation（计数问题)