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;
  inline long long read() {
    long long x = 0, f = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    return x * f;
  }
  const long long N = 4e3 + 10;
  long long n, m, ans, root, top, sta[N], s[N], siz[N], f[N][N];
  struct Tree {
    long long son[2];
  } tre[N];
  void dfs(long long x) {
    siz[x] = 1;
    for (long long i = 0; i <= 1; i++) {
      if (!tre[x].son[i]) continue;
      long long to = tre[x].son[i];
      dfs(to);
      for (long long j = min(m, siz[x]); j >= 0; j--)
        for (long long k = min(m - j, siz[to]); k >= 0; k--)
          f[x][j + k] =
              max(f[x][j + k], f[x][j] + f[to][k] + (s[to] - s[x]) * (m - k) * k);
      siz[x] += siz[to];
    }
  }
  int main() {
    n = read();
    m = read();
    for (long long i = 1; i <= n; i++) s[i] = read();
    for (long long i = 1; i <= n; i++) {
      while (top && s[sta[top]] > s[i]) tre[i].son[0] = sta[top--];
      if (top) tre[sta[top]].son[1] = i;
      sta[++top] = i;
    }
    while (top) root = sta[top--];
    dfs(root);
    printf("%lld", f[root][m]);
    return 0;
  }
  