Minesweeper

题目描述

Minesweeper ave you ever played Minesweeper? This cute little game comes with a certain operating system whose name we can't remember. The goal of the game is to find where all the mines are located within a M x N field. The game shows a number in a square which tells you how many mines there are adjacent to that square. Each square has at most eight adjacent squares. The 4 x 4 field on the left contains two mines, each represented by a ``*'' character. If we represent the same field by the hint numbers described above, we end up with the field on the right: *... .... .*.. .... *100 2210 1*10 1110

输入

The input will consist of an arbitrary number of fields. The first line of each field contains two integers n and m ( 0 < n, m$ le$100) which stand for the number of lines and columns of the field, respectively. Each of the next n lines contains exactly m characters, representing the field. Safe squares are denoted by ``.'' and mine squares by ``*,'' both without the quotes. The first field line where n = m = 0 represents the end of input and should not be processed.

输出

For each field, print the message Field #x: on a line alone, where x stands for the number of the field starting from 1. The next n lines should contain the field with the ``.'' characters replaced by the number of mines adjacent to that square. There must be an empty line between field outputs.

样例输入

4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0

样例输出

Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100

#include <iostream>
#include <cstring>

using namespace std;

int dir[8][2] = {{-1,0},{0,1},{1,0},{0,-1},{-1,-1},{-1,1},{1,1},{1,-1}};

int main()
{
    int num = 1;
    int m,n;
    cin>>m>>n;
    while(m != 0 && n != 0)
    {
        char data[m][n];
        int a[m][n];
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                cin>>data[i][j];
        memset(a,0,sizeof(a));
        int m1 = m-1;
        int n1 = n-1;
        while(m1 != 0 || n1 != 0)
        {
            if(data[m1][n1] != '*')
            {
                for(int i = 0; i<8; i++)
                {
                    int x = m1+dir[i][0];
                    int y = n1+dir[i][1];
                    if( x > -1 &&  y > -1 && y < n && x < m)
                    {
                        if(data[x][y] == '*' && data[m1][n1] == '.')
                            a[m1][n1]++;
                    }
                }
            }
            if(n1-1 != -1)
                n1--;
            else
            {
                n1 = n-1;
                m1--;
            }

        }
        cout<<"Field #"<<num<<":"<<endl;
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(data[i][j] == '*')
                    cout<<"*";
                else
                    cout<<a[i][j];
            }
            cout<<endl;
        }
        cout<<endl;
        num++;
        cin>>m>>n;
    }
    return 0;
}