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 = 105;

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 == 1 ? 0 : INF;
    if (r - l + 1 == k) return 0;
    if (dp[l][r][k] != -1) return dp[l][r][k];
    int ans = INF;
    if (k == 1)
    {
        for (int i = L; i <= R; i++)
            ans = min(ans, dfs(l, r, i) + sum[r] - sum[l - 1]);
    }
    else
    {
        for (int i = 0; i < r - l; i++)
            ans = min(ans, dfs(l, l + i, 1) + dfs(l + i + 1, r, k - 1));
    }
    return dp[l][r][k] = ans;
}

int main()
{
    while (cin >> n >> L >> R)
    {
        memset(sum, 0, sizeof sum);
        memset(dp, -1, sizeof dp);
        for (int i = 1; i <= n; i++)
        {
            cin >> a[i];
            sum[i] = sum[i - 1] + a[i];
        }
        int ans = dfs(1, n, 1);
        cout << (ans == INF ? 0 : ans) << endl;
    }
    return 0;
}