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, a[4001], ls[4001], rs[4001], siz[4001];
  long long f[4001][4001];
  int build(int l, int r) {
    if (l > r) return 0;
    int x = l;
    for (int i = l + 1; i <= r; i++)
      if (a[i] < a[x]) x = i;
    ls[x] = build(l, x - 1);
    rs[x] = build(x + 1, r);
    return x;
  }
  void dfs(int u) {
    siz[u] = 1;
    if (ls[u]) dfs(ls[u]), siz[u] += siz[ls[u]];
    if (rs[u]) dfs(rs[u]), siz[u] += siz[rs[u]];
    for (int i = 0; i <= siz[ls[u]]; i++)
      for (int j = 0; j <= siz[rs[u]]; j++)
        f[u][i + j] =
            max(f[u][i + j], f[ls[u]][i] + f[rs[u]][j] - 2ll * i * j * a[u]),
                 f[u][i + j + 1] =
                     max(f[u][i + j + 1], f[ls[u]][i] + f[rs[u]][j] +
                                              (M - 1ll) * a[u] -
                                              2ll * (i + j + i * j) * a[u]);
  }
  int main() {
    cin >> N >> M;
    for (int i = 1; i <= N; i++) cin >> a[i];
    int rt = build(1, N);
    dfs(rt);
    cout << f[rt][M] << '\n';
    return 0;
  }
  