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

 

位向量解决 

 

#include <stdio.h>#include <string.h>#define max 100char mine[max][max];int main(){    int m, n, i, j, k, count = 1;    int px[8] = {-1,-1,-1, 0, 0, 1, 1, 1};    int py[8] = {-1, 0, 1,-1, 1,-1, 0, 1};    while (scanf("%d %d", &n, &m) && (m || n))    {          memset(mine, '.', sizeof(mine));          for (i = 0; i < n; i++)              for (j = 0; j < m; j++)                  scanf(" %c", &mine[i][j]);          for (i = 0; i < n; i++)              for (j = 0; j < m; j++)              {                  int tmp = 0;                  if (mine[i][j] == '.')                     for (k = 0; k < 8; k++)                     {                                                  if (mine[i+px[k]][j+py[k]] == '*')                            tmp++;                         mine[i][j] = tmp + '0';                     }              }          if (count != 1)            printf("\n");          printf("Field #%d:\n", count++);          for (i = 0; i < n; i++)          {              for (j = 0; j < m; j++)                  printf("%c", mine[i][j]);              printf("\n");          }    }    return 0;}