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, rt, a[4005], st[4005], sz[4005], ch[4005][2];
  long long dp[4005][4005];
  void dfs(int u) {
    if (!u) return;
    dfs(ch[u][0]);
    dfs(ch[u][1]);
    sz[u] = sz[ch[u][0]] + sz[ch[u][1]] + 1;
    for (int i = 0; i <= sz[ch[u][0]]; ++i)
      for (int j = 0; j <= sz[ch[u][1]]; ++j)
        dp[u][i + j] = max(
            dp[u][i + j], dp[ch[u][0]][i] + dp[ch[u][1]][j] - 2ll * i * j * a[u]);
    for (int i = sz[u] - 1; i >= 0; --i)
      dp[u][i + 1] = max(dp[u][i + 1], dp[u][i] + 1ll * (m - i * 2 - 1) * a[u]);
  }
  int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) {
      while (st[0] && a[i] < a[st[st[0]]]) ch[i][0] = st[st[0]--];
      ch[st[st[0]]][1] = i;
      st[++st[0]] = i;
    }
    dfs(st[1]);
    printf("%lld\n", dp[st[1]][m]);
    return 0;
  }
  