uva 10189 - 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:

*10022101*101110

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:*10022101*101110Field #2:**100332001*100

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© 2001 Universidade do Brasil (UFRJ). Internal Contest Warmup 2001.

本来很简单的一道题，wa了3次，也不知然后在输入周围加了一圈空白，就ac了

#include<stdio.h>int main(){int n,m,i,j,count=1,num;char st[105][105];while(scanf("%d%d",&n,&m)!=EOF){getchar();if(n==0&&m==0) break;        for(i=1;i<=n;i++){for(j=1;j<=m;j++){scanf("%c",&st[i][j]);}   getchar();}for(i=1;i<=m+1;i++){st[n+1][i]='.';}for(i=1;i<=n+1;i++){st[i][m+1]='.';}if(count!=1)printf("\n");for(i=1;i<=n;i++){for(j=1;j<=m;j++){num=0;if(st[i][j]!='*'){if(st[i-1][j]=='*') num++;if(st[i+1][j]=='*') num++;if(st[i][j-1]=='*') num++;if(st[i][j+1]=='*') num++;if(st[i-1][j-1]=='*') num++;if(st[i-1][j+1]=='*') num++;if(st[i+1][j-1]=='*') num++;if(st[i+1][j+1]=='*') num++;st[i][j]=(char)num+'0';}}}printf("Field #%d:\n",count);count++;for(i=1;i<=n;i++){for(j=1;j<=m;j++)printf("%c",st[i][j]);printf("\n");}}return 0;}