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;
  const int c = 4005;
  long long n, m, t[c], fel[c], dp[c][c], rf[c];
  vector<int> sz[c];
  void solve(int a, int b, int c) {
    if (a > b) {
      return;
    }
    int pos = a;
    for (int i = a; i <= b; i++) {
      if (t[i] < t[pos]) {
        pos = i;
      }
    }
    sz[c].push_back(pos);
    fel[pos] = c;
    solve(a, pos - 1, pos), solve(pos + 1, b, pos);
  }
  void dfs(int a) {
    rf[a] = 1;
    for (auto x : sz[a]) {
      dfs(x);
      for (int i = rf[a]; i >= 0; i--) {
        for (int j = rf[x]; j >= 0; j--) {
          dp[a][i + j] = max(dp[a][i + j],
                             dp[a][i] + dp[x][j] + j * (m - j) * (t[x] - t[a]));
        }
      }
      rf[a] += rf[x];
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> t[i];
    }
    solve(1, n, 0);
    int st = sz[0][0];
    dfs(st);
    cout << dp[st][m] << "\n";
    return 0;
  }
  