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>
  template <typename T>
  void checkmax(T &x, T y) {
    if (x < y) x = y;
  }
  template <typename T>
  void checkmin(T &x, T y) {
    if (x > y) x = y;
  }
  int n, m, sz[4001];
  long long a[4001], dp[4001][4001];
  int Solve(int l, int r) {
    if (l > r) return 0;
    int p = std::min_element(a + l, a + r + 1) - a, u = Solve(l, p - 1),
        v = Solve(p + 1, r);
    sz[p] = sz[u] + sz[v] + 1;
    for (int i = 0; i <= sz[u]; i++)
      for (int j = 0; j <= sz[v]; j++) {
        checkmax(dp[p][i + j], dp[u][i] + dp[v][j] - 2 * a[p] * i * j);
        checkmax(dp[p][i + j + 1], dp[u][i] + dp[v][j] + a[p] * (m - 1) -
                                       2 * a[p] * (i * j + i + j));
      }
    return p;
  }
  int main(int argc, char const *argv[]) {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr), std::cout.tie(nullptr);
    std::cin >> n >> m;
    for (int i = 1; i <= n; i++) std::cin >> a[i];
    std::memset(dp, 0xcf, sizeof(dp));
    dp[0][0] = 0LL;
    std::cout << dp[Solve(1, n)][m];
    return 0;
  }
  