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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
    agc033a()
}

private data class Coordinate(val y: Int, val x: Int)

fun agc033a() {
    val (h, w) = readLine()!!.split(' ').map { it.toInt() }
    val isBlack = (1..h).map { readLine()!!.map { it == '#' }.toBooleanArray() }.toTypedArray()

    val blackIndex = isBlack.mapIndexed { indexY, booleans ->
        booleans.withIndex().filter { it.value }.map { Coordinate(indexY, it.index) }
    }.flatten()

    val queue = ArrayDeque(listOf(blackIndex))
    val visited = Array(h) { BooleanArray(w) }
    var answer = 0

    while (queue.isNotEmpty()) {
        val poll = queue.poll()
        if (poll.isEmpty()) continue

        val list = mutableListOf<Coordinate>()

        poll.forEach {
            val (y, x) = it
            visited[y][x] = true

            if ((y + 1 in 0 until h && x in 0 until w) && !visited[y + 1][x]) {
                list.add(Coordinate(y + 1, x))
            }
            if ((y - 1 in 0 until h && x in 0 until w) && !visited[y - 1][x]) {
                list.add(Coordinate(y - 1, x))
            }
            if ((y in 0 until h && x + 1 in 0 until w) && !visited[y][x + 1]) {
                list.add(Coordinate(y, x + 1))
            }
            if ((y in 0 until h && x - 1 in 0 until w) && !visited[y][x - 1]) {
                list.add(Coordinate(y, x - 1))
            }
        }

        if (list.isEmpty()) continue

        queue.add(list)

        answer++
    }

    println(answer)
}