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 JavaScript (node.js v5.12) (judge verdict: Accepted)



function Main(input) {
  input = input.split("\n");
  var H = parseInt(input[0].split(" ")[0]);
  var W = parseInt(input[0].split(" ")[1]);
  var A = new Array(H);
  var A = new Array(H+2);
  var C = new Array();
  var C1=new Array(W+2).fill(-1);
  A[0]=C1;
  for(var i=0;i<H;i++){
    var tmp=input[i+1].split("");
    for(var j=0;j<W;j++){
      if(tmp[j]=="#"){
        C.push([i+1,j+1]);
        tmp[j]=0;
      }
    }
    tmp.splice(0,0,-1);
    tmp.push(-1);
    A[i+1]=tmp;
  }
  A[H+1]=C1;
  var ans =0;
  for(var i=0;i<H*W;i++){
    if(A[C[i][0]+1][C[i][1]]=="."){
      A[C[i][0]+1][C[i][1]]=A[C[i][0]][C[i][1]]+1;
      C.push([C[i][0]+1,C[i][1]]);
      ans=Math.max(ans,A[C[i][0]][C[i][1]]+1);
    }
    if(A[C[i][0]-1][C[i][1]]=="."){
      A[C[i][0]-1][C[i][1]]=A[C[i][0]][C[i][1]]+1;
      C.push([C[i][0]-1,C[i][1]]);
      ans=Math.max(ans,A[C[i][0]][C[i][1]]+1);
    }
    if(A[C[i][0]][C[i][1]+1]=="."){
      A[C[i][0]][C[i][1]+1]=A[C[i][0]][C[i][1]]+1;
      C.push([C[i][0],C[i][1]+1]);
      ans=Math.max(ans,A[C[i][0]][C[i][1]]+1);
    }
    if(A[C[i][0]][C[i][1]-1]=="."){
      A[C[i][0]][C[i][1]-1]=A[C[i][0]][C[i][1]]+1;
      C.push([C[i][0],C[i][1]-1]);
      ans=Math.max(ans,A[C[i][0]][C[i][1]]+1);
    }
  }

  console.log('%s', ans);
  return ans;
}

function debug(n){
  var input = document.getElementById("input"+n).value;
  var ans=Main(input);
  var result="WA";
  if(ans==document.getElementById("output"+n).value.split("\n")[0]){
    result="AC";
  }
  document.getElementById("result"+n).innerHTML=result;
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));