简单粗暴好理解

解题思路:
单独加一行一列来保证每个格子的八个相邻位置都是干净的。

参考代码:

#include<iostream>
using namespace std;

int main()
{
    int m,n,sum;
    int f=1;
    while(cin>>m>>n&&m*n!=0)//保证可以重复输入且m和n都不为0
    {
        sum=0;              //用来统计每个格子旁边有几个地雷
        char a[m+1][n+1]= {0};//在输入给出的行和列的基础上多加了一行一列，从而保证边缘格子的8个空间也都是干净的，基础值全部为0
        for(int i=1; i<m+1; i++)
            for(int j=1; j<n+1; j++)
                cin>>a[i][j];
        cout<<"Field #"<<f<<":"<<endl;
        for(int i=1; i<m+1; i++)//从这里开始输出具体的值
        {
            for(int j=1; j<n+1; j++)
            {
                if(a[i][j]=='*')//若格子本身是地雷那就继续输出“*”
                    cout<<'*';
                else             //若格子不是地雷，那么就依次查看八个格子的值，简单粗暴！
                {
                    sum=0;
                    if(a[i-1][j-1]=='*'&&i-1>-1&&j-1>-1)
                        sum++;
                    if(a[i-1][j]=='*')
                        sum++;
                    if(a[i-1][j+1]=='*')
                        sum++;
                    if(a[i][j-1]=='*')
                        sum++;
                    if(a[i][j+1]=='*')
                        sum++;
                    if(a[i+1][j-1]=='*')
                        sum++;
                    if(a[i+1][j]=='*')
                        sum++;
                    if(a[i+1][j+1]=='*')
                        sum++;
                    cout<<sum;
                }
            }
            cout<<endl;
        }
        f++;                              //用来控制输出的标号，也就是“Field #x:”位置的x值
        cout<<endl;
    }
    return 0;
}