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;
  long long n, m, dsu[4069], cc[4069], dp[4069][4069], tmp[4069];
  pair<long long, long long> a[4069];
  long long fd(long long x) {
    if (dsu[x] != x) {
      dsu[x] = fd(dsu[x]);
    }
    return dsu[x];
  }
  int main() {
    long long i, j, r, u, k, l, w, z = 0;
    scanf("%lld%lld", &n, &m);
    for (i = 1; i <= n; i++) {
      scanf("%lld", &k);
      a[i] = {k, i};
    }
    sort(a + 1, a + n + 1, greater<pair<long long, long long>>());
    for (i = 1; i <= n; i++) {
      w = a[i].first;
      k = a[i].second;
      dsu[k] = k;
      cc[k] = 1;
      dp[k][1] = w * m - w;
      for (u = -1; u <= 1; u += 2) {
        if (dsu[k + u]) {
          l = fd(k + u);
          for (j = 0; j <= cc[k]; j++) {
            for (r = 0; r <= cc[l]; r++) {
              tmp[j + r] = max(tmp[j + r], dp[k][j] + dp[l][r] - w * j * r * 2);
            }
          }
          cc[k] += cc[l];
          dsu[l] = k;
          for (j = 0; j <= cc[k]; j++) {
            dp[k][j] = tmp[j];
            tmp[j] = 0;
          }
        }
      }
      z = max(z, dp[k][m]);
    }
    printf("%lld\n", z);
  }
  