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题目描述

Minesweeper Have 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<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

提示

代码如下

#include <stdio.h>
#include <string.h>
int main()
{
    char lei[120][120];
    int ci=0,n,m;
    while(~scanf("%d%d",&n,&m)&&(n||m))
    {
        memset(lei,'0',sizeof(lei));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                char x;
                scanf(" %c",&x);
                if(x=='*')
                {
                    lei[i][j]='*';
                    for(int ii=i-1; ii<=i+1; ii++)
                        for(int jj=j-1; jj<=j+1; jj++)
                            if(lei[ii][jj]!='*')lei[ii][jj]++;
                }
            }
        printf("Field #%d:\n",++ci);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)printf(j!=m?"%c":"%c\n",lei[i][j]);
        printf("\n");
    }
    return 0;
}

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