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;
  const int N = 4005;
  int pos[N][N], ls[N], rs[N], siz[N];
  int build(int l, int r) {
    if (l == r) return l;
    int x = pos[l][r];
    if (x - 1 >= l) ls[x] = build(l, x - 1);
    if (x + 1 <= r) rs[x] = build(x + 1, r);
    return x;
  }
  int n, m;
  long long a[N], dp[N][N];
  void dfs(int x) {
    siz[x] = 1;
    if (ls[x]) dfs(ls[x]), siz[x] += siz[ls[x]];
    if (rs[x]) dfs(rs[x]), siz[x] += siz[rs[x]];
    if ((!ls[x]) && (!rs[x])) return;
    if (!ls[x]) {
      for (int i = 0; i <= siz[rs[x]]; i++) {
        dp[x][i] = max(dp[x][i], dp[rs[x]][i] + (a[rs[x]] - a[x]) * i * (m - i));
        dp[x][i + 1] =
            max(dp[x][i + 1], dp[rs[x]][i] + (a[rs[x]] - a[x]) * i * (m - i));
      }
      return;
    }
    if (!rs[x]) {
      for (int i = 0; i <= siz[ls[x]]; i++) {
        dp[x][i] = max(dp[x][i], dp[ls[x]][i] + (a[ls[x]] - a[x]) * i * (m - i));
        dp[x][i + 1] =
            max(dp[x][i + 1], dp[ls[x]][i] + (a[ls[x]] - a[x]) * i * (m - i));
      }
      return;
    }
    for (int i = 0; i <= siz[ls[x]]; i++)
      for (int j = 0; j <= siz[rs[x]]; j++) {
        long long add =
            (a[ls[x]] - a[x]) * i * (m - i) + (a[rs[x]] - a[x]) * j * (m - j);
        dp[x][i + j] = max(dp[x][i + j], dp[ls[x]][i] + dp[rs[x]][j] + add);
        dp[x][i + j + 1] =
            max(dp[x][i + j + 1], dp[ls[x]][i] + dp[rs[x]][j] + add);
      }
  }
  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++) {
      pos[i][i] = i;
      for (int j = i + 1; j <= n; j++)
        pos[i][j] = a[pos[i][j - 1]] < a[j] ? pos[i][j - 1] : j;
    }
    build(1, n);
    dfs(pos[1][n]);
    printf("%lld", dp[pos[1][n]][m]);
    return 0;
  }
  