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, sz[4010];
  long long dp[4010][4010], a[4010];
  int sol(int l, int r) {
    if (l > r) return 0;
    int id = l;
    for (int i = l + 1; i <= r; i++)
      if (a[id] > a[i]) id = i;
    int ls = sol(l, id - 1), rs = sol(id + 1, r);
    sz[id] = sz[ls] + sz[rs] + 1;
    for (int i = 0; i <= sz[ls]; i++)
      for (int j = 0; j <= sz[rs]; j++)
        dp[id][i + j] =
            max(dp[id][i + j], dp[ls][i] + dp[rs][j] - a[id] * i * j * 2);
    for (int i = sz[id]; i; i--)
      dp[id][i] = max(dp[id][i], dp[id][i - 1] + a[id] * (m - 2 * i + 1));
    return id;
  }
  int main() {
    memset(dp, -0x7f, sizeof(dp));
    dp[0][0] = 0;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
    int rt = sol(1, n);
    return printf("%lld", dp[rt][m]), 0;
  }
  