description: Alice has an integer sequence a of length n and all elements are different. She will choose a subsequence of a of length m, and defines the value of a subsequence a_{b_1},a_{b_2},…,a_{b_m} as $$$∑_{i = 1}^m (m ⋅ a_{b_i}) - ∑_{i = 1}^m ∑_{j = 1}^m f(min(b_i, b_j), max(b_i, b_j)), where f(i, j) denotes \min(a_i, a_{i + 1}, \ldots, a_j)$$$.
  
  Alice wants you to help her to maximize the value of the subsequence she choose.
  
  A sequence s is a subsequence of a sequence t if s can be obtained from t by deletion of several (possibly, zero or all) elements.
  
  Input
  
  The first line contains two integers n and m (1 ≤ m ≤ n ≤ 4000).
  
  The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i < 2^{31}).
  
  Output
  
  Print the maximal value Alice can get.
  
  Examples
  
  Input
  
  
  6 4
  15 2 18 12 13 4
  
  
  Output
  
  
  100
  
  
  Input
  
  
  11 5
  9 3 7 1 8 12 10 20 15 18 5
  
  
  Output
  
  
  176
  
  
  Input
  
  
  1 1
  114514
  
  
  Output
  
  
  0
  
  
  Input
  
  
  2 1
  666 888
  
  
  Output
  
  
  0
  
  Note
  
  In the first example, Alice can choose the subsequence [15, 2, 18, 13], which has the value 4 ⋅ (15 + 2 + 18 + 13) - (15 + 2 + 2 + 2) - (2 + 2 + 2 + 2) - (2 + 2 + 18 + 12) - (2 + 2 + 12 + 13) = 100. In the second example, there are a variety of subsequences with value 176, and one of them is [9, 7, 12, 20, 18].
solution: #include <bits/stdc++.h>
  using namespace std;
  long long int a[4005];
  long long int dp[4005][4005];
  int m;
  int dfs(int l, int r) {
    if (l > r) return 0;
    long long int mi = 1e18;
    long long int root = -1;
    for (int i = l; i <= r; i++) {
      if (a[i] < mi) {
        mi = a[i];
        root = i;
      }
    }
    int root_l = dfs(l, root - 1);
    int root_r = dfs(root + 1, r);
    int sz = r - l + 1;
    for (int i = 0; i <= sz; i++) {
      int sz_l = root - l;
      int sz_r = r - root;
      dp[root][i] = -1e18;
      for (int j = max(0, i - sz_r - 1); j <= min(sz_l, i); j++) {
        dp[root][i] = max(dp[root][i], dp[root_l][j] + dp[root_r][i - j] -
                                           2ll * j * (i - j) * a[root]);
        if (i - 1 - j >= 0)
          dp[root][i] = max(dp[root][i], dp[root_l][j] + dp[root_r][i - 1 - j] -
                                             2ll * (j + 1) * (i - j) * a[root] +
                                             a[root] + 1ll * m * a[root]);
      }
    }
    return root;
  }
  int main(void) {
    int n;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    for (int i = 0; i <= n; i++) {
      dp[0][i] = -1e18;
    }
    dp[0][0] = 0;
    int root = dfs(1, n);
    printf("%lld\n", dp[root][m]);
    return 0;
  }
  