description: You are given a strictly convex polygon with n vertices.
  
  You will make k cuts that meet the following conditions: 
  
    * each cut is a segment that connects two different nonadjacent vertices; 
    * two cuts can intersect only at vertices of the polygon. 
  
  
  
  Your task is to maximize the area of the smallest region that will be formed by the polygon and those k cuts.
  
  Input
  
  The first line contains two integers n, k (3 ≤ n ≤ 200, 0 ≤ k ≤ n-3).
  
  The following n lines describe vertices of the polygon in anticlockwise direction. The i-th line contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^8) — the coordinates of the i-th vertex.
  
  It is guaranteed that the polygon is convex and that no two adjacent sides are parallel.
  
  Output
  
  Print one integer: the maximum possible area of the smallest region after making k cuts multiplied by 2.
  
  Examples
  
  Input
  
  
  8 4
  -2 -4
  2 -2
  4 2
  1 5
  0 5
  -4 4
  -5 0
  -5 -1
  
  
  Output
  
  
  11
  
  
  Input
  
  
  6 3
  2 -2
  2 -1
  1 2
  0 2
  -2 1
  -1 0
  
  
  Output
  
  
  3
  
  Note
  
  In the first example, it's optimal to make cuts between the following pairs of vertices: 
  
    * (-2, -4) and (4, 2), 
    * (-2, -4) and (1, 5), 
    * (-5, -1) and (1, 5), 
    * (-5, 0) and (0, 5). 
  
  <image> Points (-5, -1), (1, 5), (0, 5), (-5, 0) determine the smallest region with double area of 11. 
  
  In the second example, it's optimal to make cuts between the following pairs of vertices: 
  
    * (2, -1) and (0, 2), 
    * (2, -1) and (1, 0), 
    * (-1, 0) and (0, 2). 
  
  <image> Points (2, -2), (2, -1), (-1, 0) determine one of the smallest regions with double area of 3. 
solution: #include <bits/stdc++.h>
  using namespace std;
  inline long long readint() {
    long long x = 0;
    bool f = 0;
    char c = getchar();
    while (!isdigit(c) && c != '-') c = getchar();
    if (c == '-') {
      f = 1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = x * 10 + c - '0';
      c = getchar();
    }
    return f ? -x : x;
  }
  const int maxn = 200 + 5;
  int n, k, x[maxn], y[maxn];
  long long S(int a, int b, int c) {
    return abs(1ll * (x[b] - x[a]) * (y[c] - y[a]) -
               1ll * (x[c] - x[a]) * (y[b] - y[a]));
  }
  int f[maxn][maxn];
  long long g[maxn][maxn];
  int main() {
    n = readint();
    k = readint();
    for (int i = 1; i <= n; i++) {
      x[i] = readint();
      y[i] = readint();
    }
    long long l = 0, r = 8e16;
    while (l < r) {
      long long mid = l + (r - l + 1) / 2;
      for (int i = n; i > 0; i--)
        for (int j = i + 1; j <= n; j++) {
          f[i][j] = 0;
          g[i][j] = 0;
          for (int k = i + 1; k < j; k++) g[i][j] += S(j, k, k - 1);
          auto upd = [i, j](int f2, long long g2) {
            if (f2 > f[i][j] || (f2 == f[i][j] && g2 > g[i][j])) {
              f[i][j] = f2;
              g[i][j] = g2;
            }
          };
          for (int k = i + 1; k < j; k++)
            if (S(i, j, k) + g[i][k] + g[k][j] >= mid)
              upd(f[i][k] + f[k][j] + 1, 0);
            else
              upd(f[i][k] + f[k][j], S(i, j, k) + g[i][k] + g[k][j]);
        }
      if (f[1][n] >= k + 1)
        l = mid;
      else
        r = mid - 1;
    }
    printf("%lld\n", r);
    return 0;
  }
  