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;
  int n, m;
  int a[4086];
  int son[4086][2];
  int sk[4086], top;
  long long f[4086][4086], g[4086];
  int siz[4086];
  void dfs(int i) {
    f[i][0] = f[i][1] = 0, siz[i] = 1;
    for (int j = 0; j <= 1; j++) {
      int to = son[i][j];
      if (!to) continue;
      dfs(to);
      for (int k = 0; k <= min(siz[i], m); k++) {
        for (int l = 0; l <= min(siz[to], m - k); l++) {
          g[k + l] = max(g[k + l],
                         f[i][k] + f[to][l] + 1ll * (a[to] - a[i]) * l * (m - l));
        }
      }
      for (int k = 0; k <= min(siz[i] + siz[to], m); k++)
        f[i][k] = g[k], g[k] = -1e18;
      siz[i] += siz[to];
    }
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      int last = 0;
      while (top && a[sk[top]] > a[i]) last = sk[top--];
      son[i][0] = last, son[sk[top]][1] = i;
      sk[++top] = i;
    }
    memset(f, -0x3f, sizeof(f)), memset(g, -0x3f, sizeof(g));
    dfs(sk[1]);
    printf("%lld", f[sk[1]][m]);
  }
  