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 Rust (1.15.1) (judge verdict: Accepted)

use std::collections::VecDeque;

mod input {
    use std;

    pub fn read_line() -> String {
        let mut line = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().to_string()
    }

    pub fn read_vec<T: std::str::FromStr>() -> Vec<T> {
        read_line()
            .split(' ')
            .filter_map(|x| x.parse::<T>().ok())
            .collect::<Vec<T>>()
    }

    pub fn read_chars() -> Vec<char> {
        read_line().chars().collect()
    }
}

fn main() {
    let (n, m) = {
        let token = input::read_vec::<usize>();
        (token[0], token[1])
    };

    let a = (0..n).map(|_| input::read_chars()).collect::<Vec<_>>();

    let mut queue: VecDeque<(usize, usize)> = VecDeque::new();
    let mut dist: Vec<Vec<i32>> = vec![vec![1000005; m]; n];

    for i in 0..n {
        for j in 0..m {
            if a[i][j] == '#' {
                dist[i][j] = 0;
                queue.push_back((i, j));
            }
        }
    }

    while let Some((i, j)) = queue.pop_front() {
        for (ni, nj) in [(!0usize, 0usize), (1, 0), (0, !0), (0, 1)]
            .iter()
            .map(|&(di, dj)| (i.wrapping_add(di), j.wrapping_add(dj)))
            .filter(|&(ni, nj)| ni < n && nj < m)
        {
            if dist[i][j] + 1 < dist[ni][nj] {
                dist[ni][nj] = dist[i][j] + 1;
                queue.push_back((ni, nj));
            }
        }
    }

    if let Some(ans) = dist.iter().filter_map(|x| x.iter().max()).max() {
        println!("{}", ans);
    }
}