Solution of 10189 - Minesweeper

Problem Description
source:https://uva.onlinejudge.org/external/101/10189.html

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 M × N 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 4 × 4 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.



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.

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 

3 5 
**...
 ..... 
.*... 
0 0

Sample Output 

Field #1:
**100
33200
1*100

Solution:

#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include<stdio.h>
#include<stdlib.h>

using namespace std;

int main()
{
    int i,j,m,n,adjacent,set=1;
    int value[104][104];
    char place[104][104],ch;
    //freopen("input.txt","r",stdin);
    while(scanf("%d%d%c",&n,&m,&ch)==3)
    {
        if(m==0 && m==0)
            break;
        if(set!=1)
            printf("\n");
        for(i=0;i<n;i++)
            gets(place[i]);
        memset(value,0, sizeof(value));
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(place[i][j]=='*')
                {
                    value[i-1][j-1]+=1;
                    value[i-1][j]+=1;
                    value[i-1][j+1]+=1;
                    value[i][j+1]+=1;
                    value[i+1][j+1]+=1;
                    value[i+1][j]+=1;
                    value[i+1][j-1]+=1;
                    value[i][j-1]+=1;
                }
            }
        }
        printf("Field #%d:\n",set);
            set+=1;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(place[i][j]=='*')
                    printf("*");
                else
                    printf("%d",value[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}