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: Runtime Error)

#include <stdio.h>

int main(int argc, const char * argv[]) {
    
    int H = 0;
    int W = 0;
    
    scanf("%d",&H);
    scanf("%d",&W);
    
    char A[1000][1000] = {};
    for (int i = 0; i < H; i++) {
        for (int j = 0; j < W+1; j++) {
            scanf("%c",&A[i][j]);
        }
    }
    
    char B[1000][1000] = {};
    for (int i = 0; i < H+1; i++) {
        for (int j = 0; j < W+1; j++) {
            B[i][j] = A[i][j];
            printf("%c",A[i][j]);
        }
        printf("\n");
    }


    int flag = 0;
    int count = 0;
    int sharpCount = 0;
    while (flag == 0) {
        for (int i = 0; i < H; i++) {
            for (int j = 0; j < W; j++) {
                if (B[i][j] == '#') {
                    if (j-1 >= 0) {
                        A[i][j-1] = '#';
                    }
                    if (j+1 <= W) {
                        A[i][j+1] = '#';
                    }
                    if (i-1 >= 0) {
                        A[i-1][j] = '#';
                    }
                    if (i+1 <= H) {
                        A[i+1][j] = '#';
                    }
                }
            }
        }
        count++;//操作の回数
        for (int i = 0; i < H; i++) {
            for (int j = 0; j < W; j++) {
                B[i][j] = A[i][j];
                printf("%c",A[i][j]);
            }
            printf("\n");
        }
        
        sharpCount = 0;
        for (int i = 0; i < H; i++) {
            for (int j = 0; j < W; j++) {
                if (A[i][j] == '#') {
                    sharpCount++;
                }
            }
        }
        //まだ存在する
        if (sharpCount == (H*W)) {
            flag = 1;
        }
    }

    
    printf("%d\n",count);
    return 0;
}