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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {

	static final int INF = 9999999;
	static int h;
	static int w;
	static int[][] table;
	static Queue<int[]> sq;

	public static void main(String... args) {

		try (Scanner sc = new Scanner(System.in)) {
			String[] tmp = sc.nextLine().split(" ");
			h = Integer.parseInt(tmp[0]);
			w = Integer.parseInt(tmp[1]);

			table = new int[h][w];

			IntStream.range(0, h).forEach(y -> {
				String s = sc.nextLine();
				IntStream.range(0, w).forEach(x -> {
					table[y][x] = s.charAt(x) == '#' ? 0 : INF;
				});
			});

		}
		sq = new ArrayDeque<>(h * w);

		IntStream.range(0, h).forEach(y -> {
			IntStream.range(0, w).forEach(x -> {
				if (table[y][x] == 0) {
					sq.add(new int[] { y, x, 0 });
				}
			});
		});

		while (!sq.isEmpty()) {
			int[] b = sq.remove();
			erosion(b[0], b[1], b[2]);
		}
		
		int maxDis = Arrays.stream(table)
			.map(row -> Arrays.stream(row))
			.mapToInt(rs -> rs.reduce(0, (a,b) -> Math.max(a, b)))
			.reduce(0, (a,b) -> Math.max(a, b));
		
		System.out.println(maxDis);
	}

	static void erosion(int y, int x, int n) {
		if (n < table[y][x] || n == 0) {
			table[y][x] = n;

			if (y < h - 1) {
				sq.add(new int[] { y + 1, x, n + 1 });
			}
			if (y > 0) {
				sq.add(new int[] { y - 1, x, n + 1 });
			}
			if (x < w - 1) {
				sq.add(new int[] { y, x + 1, n + 1 });
			}
			if (x > 0) {
				sq.add(new int[] { y, x - 1, n + 1 });
			}
		}
	}
}