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>
  struct point {
    int x, y;
    point() : x(), y() {}
    point(int _x, int _y) : x(_x), y(_y) {}
    point operator-(const point &rhs) const {
      return point(x - rhs.x, y - rhs.y);
    }
    friend long long cross(const point &a, const point &b) {
      return 1ll * a.x * b.y - 1ll * a.y * b.x;
    }
  };
  int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int n, k;
    std::cin >> n >> k;
    ++k;
    std::vector<point> a(n);
    for (int i = 0; i < n; ++i) {
      std::cin >> a[i].x >> a[i].y;
    }
    auto check = [&](long long x) {
      std::vector<std::vector<std::pair<int, long long>>> f(
          n, std::vector<std::pair<int, long long>>(n));
      for (int i = n - 1; i >= 0; --i) {
        for (int j = i + 2; j < n; ++j) {
          for (int k = i + 1; k < j; ++k) {
            long long s = cross(a[k] - a[i], a[j] - a[i]);
            s += f[i][k].second + f[k][j].second;
            f[i][j] = std::max(
                f[i][j],
                std::pair<int, long long>(
                    f[i][k].first + f[k][j].first + (s >= x), s >= x ? 0ll : s));
          }
        }
      }
      return f[0][n - 1].first >= k;
    };
    long long l = 1, r = 1e18;
    while (l + 1 < r) {
      long long mid = (l + r) >> 1;
      if (check(mid)) {
        l = mid;
      } else {
        r = mid;
      }
    }
    std::cout << l << "\n";
  }
  