Problem:
Score : 300 points

Problem Statement

You are given a grid of squares with H horizontal rows and W vertical columns, where each square is painted white or black.
HW characters from A_{11} to A_{HW} represent the colors of the squares.
A_{ij} is # if the square at the i-th row from the top and the j-th column from the left is black, and A_{ij} is . if that square is white.

We will repeatedly perform the following operation until all the squares are black:

Every white square that shares a side with a black square, becomes black.

Find the number of operations that will be performed.
The initial grid has at least one black square.

Constraints

1 \leq H,W \leq 1000

A_{ij} is # or ..

The given grid has at least one black square.

Input

Input is given from Standard Input in the following format:

H W
A_{11}A_{12}...A_{1W}
:
A_{H1}A_{H2}...A_{HW}

Output

Print the number of operations that will be performed.

Sample Input 1

3 3
...
.#.
...

Sample Output 1

2

After one operation, all but the corners of the grid will be black. After one more operation, all the squares will be black.

Sample Input 2

6 6
..#..#
......
#..#..
......
.#....
....#.

Sample Output 2

3

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>

int main (void)
{
  int h, w, i, j, k, t, y;
  char a[1001][1001];
  int dp[1001][1001];
  int ans = 0;
  
  scanf("%d %d", &h, &w);
  
  for (i = 0; i < h; i++){
    scanf("%s", &a[i]);
  }
  
  for (i = 0; i < h; i++){
    for (j = 0; j < w; j++){
      dp[i][j] = 10000000;
    }
  }
  
  for (i = 0; i < h; i++){
    for (j = 0; j < w; j++){
      if (a[i][j] == '#'){
        dp[i][j] = 0;
        for (k = 0; k < h; k++){
          if (dp[k][j] > abs(i - k)) dp[k][j] = abs(i - k);
        }
        for (k = 0; k < w; k++){
          if (dp[i][k] > abs(j - k)) dp[i][k] = abs(j - k);
        }
        for (k = 1; k < (h*w); k++){
          if (((i+k) > h)||((j+k) > w)) break;
          if (dp[i+k][j+k] > abs(k * 2)) dp[i+k][j+k] = abs(k*2);
        }
        for (k = 1; k < (h*w); k++){
          if (((i-k) < 0)||((j+k) > w)) break;
          if (dp[i-k][j+k] > abs(k * 2)) dp[i-k][j+k] = abs(k*2);
        }
        for (k = 1; k < (h*w); k++){
          if (((i+k) > h)||((j-k) < 0)) break;
          if (dp[i+k][j-k] > abs(k * 2)) dp[i+k][j-k] = abs(k*2);
        }
        for (k = 1; k < (h*w); k++){
          if (((i-k) < 0)||((j-k) < 0)) break;
          if (dp[i-k][j-k] > abs(k * 2)) dp[i-k][j-k] = abs(k*2);
        }
      }
    }
  }
  for (i = 0; i < h; i++){
    for (j = 0; j < w; j++){
      if (dp[i][j] > ans) ans = dp[i][j];
    }
  }
  
  printf("%d", ans);
    
  return 0;
}