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: Accepted)

#include <bits/stdc++.h>
using namespace std;

int N, M;
int MAX = 1000;
vector<vector<int>>  D(MAX, vector<int>(MAX));
vector<vector<char>> S(MAX, vector<char>(MAX));
vector<int> dx = { 0, 1,  0, -1 };
vector<int> dy = { 1, 0, -1,  0 };
queue<pair<int, int>> Q;

int main() {
	cin >> N >> M;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++) {
			cin >> S.at(i).at(j);
			if (S.at(i).at(j) == '#') {
				Q.push({ i,j });
			}
		}
	}

	int ans = 0;
	while (!Q.empty()) {
		int x, y;
		tie(x, y) = Q.front();
		Q.pop();
		int d = D.at(x).at(y);
		ans = d;
		for (int k = 0; k < 4; k++) {
			int px = x + dx.at(k);
			int py = y + dy.at(k);
			if (px < 0 || px >= N || py < 0 || py >= M) continue;
			if (S.at(px).at(py) == '.') {
				Q.push({ px,py });
				S.at(px).at(py) = '#';
				D.at(px).at(py) = d + 1;
			}
		}
	}

	cout << ans << endl;
	return 0;
}