1.acwing 282石子合并问题
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 int n; 5 const int N = 310; 6 int s[N]; 7 int f[N][N]; 8 9 int main () 10 { 11 scanf("%d", &n); 12 for (int i = 1; i <= n; i ++ ) scanf("%d", &s[i]); 13 14 for (int i = 1; i <= n; i ++ ) s[i] += s[i - 1]; 15 16 //区间dp套路:从小到大枚举区间长度和区间起点,在枚举我们的决策 17 //如果len == 1,表示一堆,不需要合并,代价为0; 18 for (int len = 2; len <= n; len ++ )//按照长度从小到大枚举区间长度 19 for (int i = 1; i + len - 1 <= n; i ++ ) //枚举所有区间起点 20 { 21 int l = i, r = i + len - 1; //表示该区间的起点和终点 22 f[l][r] = 1e8; 23 for (int k = l; k < r; k ++ ) //表示该区间的分割点 24 f[l][r] = min(f[l][r],f[l][k] + f[k + 1][r] + s[r] - s[l - 1]); 25 } 26 27 cout << f[1][n] << endl; 28 return 0; 29 30 }View Code
标签:std,include,int,区间,main,dp From: https://www.cnblogs.com/rw666/p/17846200.html