[UVA] 10189 - Problem B: Minesweeper

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

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#include<stdio.h>
#include<string.h>
char s[105][105];
int main(){
    int i,j,n,m,C=0;
    while(scanf("%d%d",&n,&m)!=EOF && n+m){
        memset(s,0,sizeof(s));
        for(i=1;i<=n;i++)
            scanf("%s",&s[i][1]);
        for(i=1;i<=n;i++)
        for(j=1;j<=m;j++){
            if(s[i][j]!='*'){
                int add=0;
                if(s[i+1][j]=='*') add++;
                if(s[i][j+1]=='*') add++;
                if(s[i-1][j]=='*') add++;
                if(s[i][j-1]=='*') add++;
                if(s[i+1][j+1]=='*') add++;
                if(s[i+1][j-1]=='*') add++;
                if(s[i-1][j+1]=='*') add++;
                if(s[i-1][j-1]=='*') add++;
                s[i][j]=add+'0';
            }
        }
        if(C) puts("");
        printf("Field #%d:\n",++C);
        for(i=1;i<=n;i++)
            puts(&s[i][1]);
    }
}