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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int H, W;
    cin >> H >> W;
    map<pair<int, int>, int> data;
    queue<pair<int, int>> que;

    vector<pair<int, int>> diff;
    diff.emplace_back(0,1);
    diff.emplace_back(1,0);
    diff.emplace_back(0,-1);
    diff.emplace_back(-1,0);

    for(int i=0; i<H; i++)
    {
        for(int j=0; j<W; j++)
        {
            char temp;
            cin >> temp;
            if(temp == '#')
            {
                // 黒マスは0以上(初期値: 0)
                data[make_pair(i, j)] = 0;
                que.emplace(i, j);
            }
            else
            {
                // 白マスは-1
                data[make_pair(i, j)] = -1;
            }
        }
    }

    while(!que.empty())
    {
        auto cur = que.front();
        que.pop();

        // 現在のカーソル位置の上下左右を新たにカーソル位置とする
        for(auto it=diff.begin(); it!=diff.end(); it++)
        {
            // 境界値チェック
            int nx = cur.first + it->first;
            int ny = cur.second + it->second;
            if(nx < 0 || nx >=H || ny < 0 || ny >= W)
                continue;

            if(data.at(make_pair(nx, ny)) == -1)
            {
                data.at(make_pair(nx, ny)) = data.at(make_pair(cur.first, cur.second))++;
                que.emplace(nx, ny);
            }
        }
    }

    int max = 0;
    for(auto it : data)
    {
        if(max < it.second)
            max = it.second;
    }

    cout << max << endl;

    return 0;
}