UVA 10189

Problem B: Minesweeper

The Problem

Have you ever played Minesweeper? It's a cute little game which comes within a certain Operating System which name we can't really remember. Well, the goal of the game is to find where are all the mines within a MxN field. To help you, the game shows a number in a square which tells you how many mines there are adjacent to that square. For instance, supose the following 4x4 field with 2 mines (which are represented by an * character):

*...
....
.*..
....

If we would represent the same field placing the hint numbers described above, we would end up with:

*100
2210
1*10
1110

As you may have already noticed, each square may have at most 8 adjacent squares.

The Input

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 stands for the number of lines and columns of the field respectively. The next n lines contains exactly m characters and represent the field. Each safe square is represented by an "." character (without the quotes) and each mine square is represented by an "*" character (also without the quotes). The first field line where n = m = 0 represents the end of input and should not be processed.

The Output

For each field, you must print the following message in a line alone:

Field #x:

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 adjacent mines to that square. There must be an empty line between field outputs.

Sample Input

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

Sample Output

Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100

	这题我表示真的很无奈 ，先是没 注释掉freopen TLE ， 然后无限次WA ， 竟然是因为我定义了两次m , n

 典型的水题 ， 我的思路是找不是*的点，统计它周围八个点*的个数 ；还有一种思路是扫一遍 ， 找到一个* ， 把它赋成 - 1 ， 周围8个点+ 1。

 注意存的时候要多开一圈数组 ，不然会数组越界

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;
char a[105][105];
int b[105][105];
int m , n;
int solve(int i , int j)
{
    int num = 0;
    if(i != 0)
    {
        if(j != 0)
        {
            if(a[i - 1][j - 1] == '*')num++;
        }
        if(a[i - 1][j] == '*')num++;
        if(j != n - 1)
        {
            if(a[i - 1][j + 1] == '*')num++;
        }
    }
    if(j != 0)
    {
        if(a[i][j - 1] == '*')num++;
    }
    if(j != n - 1)
    {
        if(a[i][j + 1] == '*')num++;
    }
    if(i != m - 1)
    {
        if(j != 0)
        {
            if(a[i + 1][j - 1] == '*')num++;
        }
        if(a[i + 1][j] == '*')num++;
        if(j != n - 1)
        {
            if(a[i + 1][j + 1] == '*')num++;
        }
    }
    return num;
}
int main()
{
    int Case = 1 ;
    //freopen("in.txt" , "r" , stdin);
    while(cin >> m >> n && (m || n))
    {
        for(int  i = 0 ;  i < m ; i++)
        {
            for(int j = 0  ; j < n ; j++)
            {
                cin >> a[i][j];
            }
        }
        for(int i = 0 ; i < m ; i++)
        {
            for(int  j = 0 ; j < n ; j++)
            {
                if(a[i][j] == '*')b[i][j] = -1;
                else
                {
                    b[i][j] = solve(i , j);
                }
            }
        }
        if(Case != 1)cout << endl;
        printf("Field #%d:\n" , Case);
        Case++;
        for(int i = 0 ; i < m ; i++)
        {
            for(int  j = 0 ;  j < n ; j++)
            {
                if(b[i][j] == -1)cout << '*';
                else cout << b[i][j];
            }
            cout << endl;
        }
    }
    return 0;
}
/*
Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100
*/