HDU5135Little Zu Chongzhi's Triangles(状态压缩DP)
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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5135
题意:
给出n根木棍,随意组成几个三角形,问组合的面积最大是多少;
思路:
用状态压缩,首先把可以组成的三角形的合法情况全都放进vector里,然后枚举。
#include<bits/stdc++.h>using namespace std;int s[12];double dp[1<<12];vector<int> v;double cul(int a, int b, int c){ if(a + b <= c) return 0.0; double p = (a + b + c) * 0.5; return sqrt(p*(p-a)*(p-b)*(p-c));}int main(){ int n; while(scanf("%d", &n) != EOF && n) { v.clear(); memset(dp, 0, sizeof(dp)); for(int i = 0; i < n; i++) scanf("%d", &s[i]); sort(s, s+n); for(int i = 0; i < n; i++) for(int j = i+1; j < n; j++) for(int k = j+1; k < n; k++) { int st = (1<<i) | (1<<j) | (1<<k); dp[st] = cul(s[i], s[j], s[k]); if(s[i] + s[j] > s[k]) v.push_back(st); } for(int i = 0; i < (1 << n); i++) for(int j = 0; j < v.size(); j++) { if(i & v[j]) continue; dp[i|v[j]] = max(dp[i|v[j]], dp[i] + dp[v[j]]); } printf("%.2f\n", dp[(1<<n)-1]); } return 0;}
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