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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <algorithm>
#include <vector>
#include<cmath>
#include<numeric>

using namespace std;

int main() {
	int H, W;
	cin >> H >> W;
	vector<vector<char>> A(H, vector<char>(W));
//	vector<vector<int>> Memo(H, vector<int>(W));

	vector<vector<int>> black(2, vector<int>(H * W));
	vector<vector<int>> white(2, vector<int>(H * W));

	int k = 0;
	int s = 0;
	int maxtime = 0;

	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {

			scanf("%c", &A[i][j]);

			//			cin >> A[i][j];
			if (A[i][j] == '#') {
				black[0][k] = i;
				black[1][k] = j;
				k++;
			}
			else {
				white[0][s] = i;
				white[1][s] = j;
				s++;
			}
		}
	}


	for (int i = 0; i < s; i++) {
		
		int minDis = H * W;

		for (int j = 0; j < k; j++) {
			
			minDis = min(minDis, abs(black[0][j] - white[0][i]) + abs(black[1][j] - white[1][i]));


		}

		maxtime = max(maxtime, minDis);
//		cout << maxtime;
	}
/*	for (int i = 0; i < H; i++) {
		for (int j = 0; j < W; j++) {
			if (A[i][j] == '#') {
			}
			else {
				int minDis = H * W;
				int dis;
				for (int m = 0; m < k; m++) {
					dis = abs(black[0][m] - i) + abs(black[1][m] - j);
					minDis = min(minDis, dis);
				}
			maxtime = max(maxtime, minDis);
			}
		}
	}
*/
	cout << maxtime;

	return 0;
}