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>
  int n, m, ls[4005], rs[4005];
  long long a[4005], siz[4005], dp[4005][4005];
  std::vector<int> st;
  void cmax(long long &A, long long B) { A = std::max(A, B); }
  void dfs(int now) {
    siz[now] = 1;
    if (ls[now]) dfs(ls[now]), siz[now] += siz[ls[now]];
    if (rs[now]) dfs(rs[now]), siz[now] += siz[rs[now]];
    if (ls[now] && rs[now]) {
      for (int i = 0; i <= siz[ls[now]]; ++i)
        for (int j = 0; j <= siz[rs[now]]; ++j) {
          cmax(dp[now][i + j],
               dp[ls[now]][i] + dp[rs[now]][j] - 2 * a[now] * i * j);
          cmax(dp[now][i + j + 1], dp[ls[now]][i] + dp[rs[now]][j] +
                                       a[now] * (m - 1 - 2 * (i + j)) -
                                       2 * a[now] * i * j);
        }
    } else if (ls[now]) {
      for (int i = 0; i <= siz[ls[now]]; ++i) {
        cmax(dp[now][i], dp[ls[now]][i]);
        cmax(dp[now][i + 1], dp[ls[now]][i] + a[now] * (m - 1 - 2 * i));
      }
    } else if (rs[now]) {
      for (int i = 0; i <= siz[rs[now]]; ++i) {
        cmax(dp[now][i], dp[rs[now]][i]);
        cmax(dp[now][i + 1], dp[rs[now]][i] + a[now] * (m - 1 - 2 * i));
      }
    } else {
      dp[now][0] = 0;
      dp[now][1] = a[now] * (m - 1);
    }
  }
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n >> m;
    st.push_back(0);
    for (int i = 1; i <= n; ++i) {
      std::cin >> a[i];
      while (!st.empty() && a[st.back()] > a[i]) ls[i] = st.back(), st.pop_back();
      rs[st.back()] = i;
      st.push_back(i);
    }
    memset(dp, 0xf0, sizeof(dp));
    dfs(rs[0]);
    std::cout << dp[rs[0]][m] << std::endl;
    return 0;
  }
  