hihocoder1636 Pangu and Stones

思路：

区间dp。dp[l][r][k]表示把区间[l, r]的石子合并成k堆所需要的最小代价。

实现：

 #include <iostream>
 #include <cstring>
 using namespace std;

 const int INF = 0x3f3f3f3f;
 const int N = ;

 int a[N], sum[N], dp[N][N][N];
 int n, L, R;

 int dfs(int l, int r, int k)
 {
     if (l == r) return k ==  ?  : INF;
     if (r - l +  == k) return ;
     if (dp[l][r][k] != -) return dp[l][r][k];
     int ans = INF;
     if (k == )
     {
         for (int i = L; i <= R; i++)
             ans = min(ans, dfs(l, r, i) + sum[r] - sum[l - ]);
     }
     else
     {
         for (int i = ; i < r - l; i++)
             ans = min(ans, dfs(l, l + i, ) + dfs(l + i + , r, k - ));
     }
     return dp[l][r][k] = ans;
 }

 int main()
 {
     while (cin >> n >> L >> R)
     {
         memset(sum, , sizeof sum);
         memset(dp, -, sizeof dp);
         for (int i = ; i <= n; i++)
         {
             cin >> a[i];
             sum[i] = sum[i - ] + a[i];
         }
         int ans = dfs(, n, );
         cout << (ans == INF ?  : ans) << endl;
     }
     return ;
 }