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 Scala (2.11.7) (judge verdict: Time Limit Exceeded)

import scala.collection.mutable

object Main extends App {
  val in   = new java.util.Scanner(System.in)
  val h, w = in.nextInt
  val t    = List.fill(h)(in.next)

  // 幅優先探索を行う

  val queue   = mutable.Queue[(Int, Int)]()
  val visited = mutable.Map.empty[(Int, Int), Int]

  // 最初に与えられている黒マスをキューに追加する
  t.zipWithIndex.foreach(r => r._1.zipWithIndex.filter(_._1 == '#').foreach(v => queue += ((r._2, v._2))))

  // 黒マスは訪問済みの印を付ける
  queue.foreach(v => visited((v._1, v._2)) = 0)

  // 塗りつぶすまでの回数を数える
  while (queue.nonEmpty) {
    val v = queue.dequeue()
    val nextTo = Seq((v._1 - 1, v._2), (v._1, v._2 + 1), (v._1 + 1, v._2), (v._1, v._2 - 1))
      .filter(v => v._1 >= 0 && v._1 < h && v._2 >= 0 && v._2 < w)
      .filterNot(visited.contains)
    queue ++= nextTo
    nextTo.foreach(nv => visited((nv._1, nv._2)) = visited((v._1, v._2)) + 1)
  }
  println(visited.maxBy(_._2)._2)
}