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;
  template <typename T>
  void chkMax(T &x, T y) {
    if (y > x) x = y;
  }
  template <typename T>
  void chkMin(T &x, T y) {
    if (y < x) x = y;
  }
  template <typename T>
  void inline read(T &x) {
    int f = 1;
    x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
  }
  const int N = 4005;
  int n, m, L[N], R[N], s[N], top, sz[N], rt;
  long long a[N], f[N][N];
  void dfs(int u) {
    if (!u) return;
    sz[u] = 1;
    int ls = L[u], rs = R[u];
    dfs(ls);
    sz[u] += sz[ls];
    dfs(rs);
    sz[u] += sz[rs];
    for (int x = 0; x <= sz[ls]; x++) {
      for (int y = 0; y <= sz[rs]; y++) {
        chkMax(f[u][x + y], f[ls][x] + f[rs][y] - 2 * a[u] * x * y);
        chkMax(f[u][x + y + 1],
               f[ls][x] + f[rs][y] + (m - 1) * a[u] - 2 * a[u] * (x * y + x + y));
      }
    }
  }
  int main() {
    memset(f, 0xcf, sizeof f);
    f[0][0] = 0;
    read(n), read(m);
    for (int i = 1; i <= n; i++) read(a[i]);
    for (int i = 1; i <= n; i++) {
      while (top && a[s[top]] > a[i]) L[i] = s[top--];
      if (top) R[s[top]] = i;
      s[++top] = i;
    }
    rt = s[1];
    dfs(rt);
    printf("%lld\n", f[rt][m]);
    return 0;
  }
  