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.

 

 

童年回忆啊。。。利用二维组统计也不难啊。。。

#include
     
      #define MAX_N 100#define MAX_M 100#define MINE -100int main(){  int n, m, field_num = 0;  char square;  int k,l,i,j;  while(scanf("%d%d",&n,&m)!=EOF&&!(n==0&&m==0))  {    getchar();    if(field_num) printf("\n");    int field[MAX_N+5][MAX_M+5]={
   0};    for(i=1;i<=n;i++)    {      for(j=1;j<=m;j++)      {        square=getchar();        if(square=='*')        {          field[i][j]=MINE;          for(k=-1;k<=1;k++)            for(l=-1;l<=1;l++)              field[i+k][j+l]++;        }       }      getchar();    }    printf("Field #%d:\n",++field_num);    for(i=1;i<=n;i++)    {      for(j=1;j<=m;j++)        if(field[i][j]<0) printf("*");        else printf("%d",field[i][j]);      printf("\n");    }  }  return 0;}