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.io.{StdIn => in}
import java.io._;

object Main extends App {
  val in = new BufferedReader(new InputStreamReader(System.in))
  val Array(h, w) = in.readLine.split(" ").map(_.toInt)
  val n = h * w

  val arr = Iterator.continually(in.readLine)
    .take(h)
    .map(_.map(ch => if(ch == '#') true else false).toArray)
    .toArray

  def bfs(grid:Array[Array[Boolean]]):Int = {
    val dist = Array.ofDim[Int](h, w)
    var distMax = 0
    val dx = Array( 0, 1, 0,-1)
    val dy = Array( 1, 0,-1, 0)
    val que = collection.mutable.Queue.empty[(Int,Int)]
    for {
      i <- 0 until h
      j <- 0 until w
      if grid(i)(j)
    } que += Tuple2(i, j)
  
    while(que.nonEmpty) {
      val(y,x) = que.dequeue
  
      for(i <- 0 until 4) {
        val ny = y + dy(i)
        val nx = x + dx(i)
        if(0 <= ny && ny < h && 0 <= nx && nx < w && !grid(ny)(nx)) {
          val d = dist(y)(x) + 1
          dist(ny)(nx) = d
          grid(ny)(nx) = true
          que += Tuple2(ny, nx)
          if(distMax < d) distMax = d
        }
      }
    }
    distMax
  }

  val ans = bfs(arr)
  println(ans)
}