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 = 4010;
  int n, m, a[N], rt, tot, L[N], siz[N], R[N], ls[N], rs[N];
  long long val[N], dp[N][N];
  inline void build(int &p, int l, int r) {
    if (l > r) return;
    p = ++tot;
    int pos = l;
    for (int i = l; i <= r; ++i)
      if (a[i] < a[pos]) pos = i;
    val[p] = a[pos];
    build(ls[p], l, pos - 1);
    build(rs[p], pos + 1, r);
    dp[p][0] = 0;
    dp[p][1] = m * val[p] - val[p];
    siz[p] = 1;
    for (int i = siz[p]; i >= 0; --i)
      for (int j = 0; j <= siz[ls[p]]; ++j)
        dp[p][i + j] =
            max(dp[p][i + j], dp[p][i] + dp[ls[p]][j] - 2ll * i * j * val[p]);
    siz[p] += siz[ls[p]];
    for (int i = siz[p]; i >= 0; --i)
      for (int j = 0; j <= siz[rs[p]]; ++j)
        dp[p][i + j] =
            max(dp[p][i + j], dp[p][i] + dp[rs[p]][j] - 2ll * i * j * val[p]);
    siz[p] += siz[rs[p]];
  }
  int main() {
    memset(dp, -0x3f, sizeof(dp));
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    build(rt, 1, n);
    printf("%lld\n", dp[1][m]);
    return 0;
  }
  