UVA 12123 - Magnetic Train Tracks（计数问题）

题目链接：12123 - Magnetic Train Tracks

题意：给定n个点，求有几个锐角三角形。

思路：和UVA 11529是同类的题，枚举一个做原点，然后剩下点根据这个原点进行极角排序，然后利用two pointer去遍历一遍，找出角度小于90度的锐角，然后扣掉这些得到钝角三角形的个数，然后在用总情况去扣掉钝角就是锐角或直角

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 = 1205;
int n;
struct Point {
	double x, y;
	void read() {
		scanf("%lf%lf", &x, &y);
 	}
} p[N];

double r[N * 2];
int C(int n, int m) {
	if (n < m) return 0;
	int 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);
}

int solve(int num) {
	int tn = 0;
	for (int i = 0; i < n; i++) {
		if (i == num) continue;
		r[tn++] = cal(p[num], p[i]);
 	}
 	sort(r, r + tn);
 	for (int i = 0; i < tn; i++)
 		r[tn + i] = r[i] + 2 * pi;
	int j = 1;
	int ans = 0;
	for (int i = 0; i < tn; i++) {
		while (fabs(r[j] - r[i]) - pi / 2 <= -eps) j++;
		ans += j - i - 1;
 	}
 	return C(tn, 2) - ans;
}

int main() {
	int cas = 0;
	while (~scanf("%d", &n) && n) {
		for (int i = 0; i < n; i++)
			p[i].read();
		int ans = 0;
		for (int i = 0; i < n; i++)
			ans += solve(i);
		printf("Scenario %d:\n", ++cas);
		printf("There are %d sites for making valid tracks\n", C(n, 3) - ans);
 	}
	return 0;
}

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