hdu5135 Little Zu Chongzhi's Triangles --状压dp

来源：互联网发布：it大厦编辑：程序博客网时间：2024/05/07 21:03

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=5135

题意：n根木棒，组成若干三角形，求最大面积和。

分析：把所有木棒升序排序，可以组成三角形所有的的组合利用位运算压缩放进vector里。然后一个个找就行。

#define _CRT_SECURE_NO_DEPRECATE#include<iostream>#include<vector>#include<cstring>#include<queue>#include<stack>#include<algorithm>#include<cmath>#include<string>#include<stdio.h>#define INF 99999999#define eps 0.0001using namespace std;int n;int v[13];double dp[1 << 12];vector<int> vec;double getArea(int a, int b, int c){if (a + b <= c)return 0.0;double p = 0.5*(a + b + c);//注意是doublereturn sqrt(p*(p - a)*(p - b)*(p - c));}int main(){while (scanf("%d", &n) && n){vec.clear();memset(dp, 0, sizeof(dp));for (int i = 0; i < n; i++)scanf("%d", &v[i]);sort(v, v + n);//注意排序for (int i = 0; i < n; i++){for (int j = i + 1; j < n; j++){for (int k = j + 1; k < n; k++){int index = (1 << i) | (1 << j) | (1 << k);dp[index] = getArea(v[i], v[j], v[k]);if (v[i] + v[j]>v[k])vec.push_back(index);}}}for (int i = 0; i < (1 << n); i++){for (int j = 0; j < vec.size(); j++){if (i&vec[j])continue;dp[i | vec[j]] = max(dp[i | vec[j]], dp[i] + dp[vec[j]]);}}printf("%.2lf\n", dp[(1 << n) - 1]);}return 0;}

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