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;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 4003;
  const long long INF = 1000000009000000009;
  int n, m;
  long long a[N];
  int mini[N][N];
  long long dp[N][N];
  void dfs(int l, int r) {
    if (l > r) return;
    int x = mini[l][r];
    dfs(l, x - 1);
    dfs(x + 1, r);
    int hl = mini[l][x - 1];
    int hr = mini[x + 1][r];
    int ql = (x - l);
    int qr = (r - x);
    for (int i = 0; i <= ql; ++i) {
      for (int j = 0; j <= qr; ++j) {
        long long s = dp[hl][i] + dp[hr][j];
        s -= i * j * a[x] * 2;
        dp[x][i + j] = max(dp[x][i + j], s);
        s += m * a[x];
        s += i * j * a[x] * 2;
        s -= (i + 1) * (j + 1) * a[x] * 2;
        s += a[x];
        dp[x][i + j + 1] = max(dp[x][i + j + 1], s);
      }
    }
  }
  void solv() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
      mini[i][i] = i;
      for (int j = i + 1; j <= n; ++j) {
        if (a[j] < a[mini[i][j - 1]])
          mini[i][j] = j;
        else
          mini[i][j] = mini[i][j - 1];
      }
    }
    for (int x = 0; x <= n; ++x) {
      for (int i = 0; i <= n; ++i) {
        dp[x][i] = -INF;
      }
    }
    dp[0][0] = 0;
    dfs(1, n);
    cout << dp[mini[1][n]][m] << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 1;
    while (tt--) solv();
    return 0;
  }
  