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 i64 = long long;
  const int mxn = 4005;
  int n, m, arr[mxn];
  int ch[mxn][2];
  int rt;
  int build(int l, int r) {
    if (l > r) return 0;
    int mid = 0;
    for (int i = l; i <= r; ++i) {
      if (mid == 0 || arr[i] < arr[mid]) {
        mid = i;
      }
    }
    ch[mid][0] = build(l, mid - 1);
    ch[mid][1] = build(mid + 1, r);
    return mid;
  }
  int siz[mxn];
  i64 dp[mxn][mxn];
  void dfs(int u) {
    if (!u) return;
    dfs(ch[u][0]);
    dfs(ch[u][1]);
    siz[u] = 1 + siz[ch[u][0]] + siz[ch[u][1]];
    memset(dp[u], -0x3f, sizeof(dp[u]));
    for (int a = 0; a < (int)(std::min(siz[ch[u][0]], m) + 1); ++a) {
      for (int b = 0; b < (int)(std::min(siz[ch[u][1]], m) + 1); ++b) {
        dp[u][a + b] = std::max(dp[u][a + b], dp[ch[u][0]][a] + dp[ch[u][1]][b] -
                                                  2LL * a * b * arr[u]);
      }
    }
    for (int a = std::min(m, siz[u] - 1); a >= 0; --a) {
      dp[u][a + 1] = std::max(
          dp[u][a + 1], dp[u][a] + 1LL * m * arr[u] - 2LL * a * arr[u] - arr[u]);
    }
  }
  int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d", &arr[i]);
    rt = build(1, n);
    memset(dp[0], -0x3f, sizeof(dp[0]));
    dp[0][0] = 0;
    dfs(rt);
    printf("%lld\n", dp[rt][m]);
    return 0;
  }
  