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 read();
  const int N = 4005;
  int n, m, st[N], top, ch[N][2], siz[N];
  long long dp[N][N], a[N];
  void dfs(int u) {
    if (!u) return;
    int lc = ch[u][0], rc = ch[u][1];
    dfs(lc);
    dfs(rc);
    siz[u] = siz[lc] + siz[rc] + 1;
    for (int i = (0); i <= (siz[lc]); ++i)
      for (int j = (0); j <= (siz[rc]); ++j) {
        dp[u][i + j] =
            max(dp[u][i + j], dp[lc][i] + dp[rc][j] - 2ll * i * j * a[u]);
        dp[u][i + j + 1] = max(
            dp[u][i + j + 1], dp[lc][i] + dp[rc][j] + m * a[u] -
                                  2ll * i * j * a[u] - (i + j) * a[u] * 2 - a[u]);
      }
  }
  int main() {
    cin >> n >> m;
    for (int i = (1); i <= (n); ++i) cin >> a[i];
    for (int i = (1); i <= (n); ++i) {
      while (top && a[st[top]] > a[i]) ch[i][0] = st[top--];
      if (top) ch[st[top]][1] = i;
      st[++top] = i;
    }
    dfs(st[1]);
    cout << dp[st[1]][m] << endl;
    return 0;
  }
  