UVA 12123 - Magnetic Train Tracks（计数问题）

来源：互联网发布：淘宝赣南脐橙编辑：程序博客网时间：2024/05/03 02:15

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;}

1 0