description: Mr. Chanek's city can be represented as a plane. He wants to build a housing complex in the city.
  
  There are some telephone poles on the plane, which is represented by a grid a of size (n + 1) × (m + 1). There is a telephone pole at (x, y) if a_{x, y} = 1.
  
  For each point (x, y), define S(x, y) as the square of the Euclidean distance between the nearest pole and (x, y). Formally, the square of the Euclidean distance between two points (x_1, y_1) and (x_2, y_2) is (x_2 - x_1)^2 + (y_2 - y_1)^2.
  
  To optimize the building plan, the project supervisor asks you the sum of all S(x, y) for each 0 ≤ x ≤ n and 0 ≤ y ≤ m. Help him by finding the value of ∑_{x=0}^{n} {∑_{y=0}^{m} {S(x, y)}}.
  
  Input
  
  The first line contains two integers n and m (0 ≤ n, m < 2000) — the size of the grid.
  
  Then (n + 1) lines follow, each containing (m + 1) integers a_{i, j} (0 ≤ a_{i, j} ≤ 1) — the grid denoting the positions of telephone poles in the plane. There is at least one telephone pole in the given grid.
  
  Output
  
  Output an integer denoting the value of ∑_{x=0}^{n} {∑_{y=0}^{m} {S(x, y)}}.
  
  Examples
  
  Input
  
  
  2 2
  101
  000
  000
  
  
  Output
  
  
  18
  
  
  Input
  
  
  5 4
  10010
  00000
  01000
  00001
  00100
  00010
  
  
  Output
  
  
  36
  
  Note
  
  <image>
  
  In the first example, the nearest telephone pole for the points (0,0), (1,0), (2,0), (0,1), (1,1), and (2,1) is at (0, 0). While the nearest telephone pole for the points (0, 2), (1,2), and (2,2) is at (0, 2). Thus, ∑_{x=0}^{n} {∑_{y=0}^{m} {S(x, y)}} = (0 + 1 + 4) + (1 + 2 + 5) + (0 + 1 + 4) = 18.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long n, m, ans;
  long long dis[2010][2010], c[2010][2010];
  char s[2010][2010];
  inline void solve(long long row, long long tl, long long tr, long long l,
                    long long r) {
    if (l > r) return;
    long long mid = (l + r) >> 1;
    dis[row][mid] = 1000000000;
    long long t = tl;
    for (long long i = tl; i <= tr; i++) {
      long long val = c[row][i] + (i - mid) * (i - mid);
      if (val < dis[row][mid]) {
        dis[row][mid] = val;
        t = i;
      }
    }
    solve(row, tl, t, l, mid - 1);
    solve(row, t, tr, mid + 1, r);
  }
  signed main() {
    scanf("%lld%lld", &n, &m);
    for (long long i = 0; i <= n; i++) scanf("%s", s[i]);
    for (long long i = 0; i <= n; i++) {
      for (long long j = 0; j <= m; j++) c[i][j] = 1000000000;
    }
    for (long long j = 0; j <= m; j++) {
      long long res = 10000;
      for (long long i = 0; i <= n; i++) {
        if (s[i][j] == '1') res = i;
        c[i][j] = min(c[i][j], (i - res) * (i - res));
      }
      res = 10000;
      for (long long i = n; ~i; i--) {
        if (s[i][j] == '1') res = i;
        c[i][j] = min(c[i][j], (i - res) * (i - res));
      }
    }
    for (long long i = 0; i <= n; i++) {
      solve(i, 0, m, 0, m);
      for (long long j = 0; j <= m; j++) ans += dis[i][j];
    }
    printf("%lld\n", ans);
    return 0;
  }
  