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 MAXN = 4005;
  const long long INF = 1ll << 60;
  int n, m, a[MAXN];
  long long f[MAXN][MAXN];
  int Dfs(int l, int r) {
    if (l > r) return 0;
    if (l == r) {
      f[l][0] = 0;
      f[l][1] = (m - 1ll) * a[l];
      return l;
    }
    int w = l;
    for (int i = l + 1; i <= r; i++)
      if (a[i] < a[w]) w = i;
    long long *f1 = f[Dfs(l, w - 1)], *f2 = f[Dfs(w + 1, r)];
    for (int i = 1; i <= r - l + 1; i++) f[w][i] = -INF;
    int l1 = w - l, l2 = r - w;
    for (int i = 0; i <= l1; i++)
      for (int j = 0; j <= l2; j++)
        f[w][i + j] = max(f[w][i + j], f1[i] + f2[j] - 2ll * i * j * a[w]);
    for (int i = r - l + 1; i; i--)
      f[w][i] = max(f[w][i], f[w][i - 1] + a[w] * (m + 1ll - i * 2));
    return w;
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", a + i);
    printf("%lld\n", f[Dfs(1, n)][m]);
    return 0;
  }
  