Problem B: Minesweeper (Uva 10189)

 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

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

int main(int argc, const char *argv[])
{
  int n, m;
  int i, j;
  int p, q;
  char mine[100][100];
  int x1, x2, y1, y2;
  int count;
  int num = 0;

  while (scanf("%d %d", &n, &m) && (n != 0 || m != 0)) {
    if (num != 0) printf("\n");
    num++;
    getchar();
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
    scanf("%c", &mine[i][j]);
      }
      getchar();
    }

    printf("Field #%d:\n", num);

    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
    if (mine[i][j] == '*') {
      printf("*"); 
      continue;
    }

    if (j - 1 < 0) x1 = 0;
    else x1 = j - 1;

    if (j + 1 > m-1) x2 = m - 1;
    else x2 = j + 1;

    if (i - 1 < 0) y1 = 0;
    else y1 = i - 1;

    if (i + 1 > n-1) y2 = n - 1;
    else y2 = i + 1;

    count = 0;
    for (p = y1; p <= y2; p++) {
      for (q = x1; q <= x2; q++) {
        if (p == i && q == j) continue;
        if (mine[p][q] == '*')
          count++;
      }
    }
    printf("%d", count);
      } 
      printf("\n");
    }
  }
    return 0;
}

总结：

　　问题很简单，但是在编写代码时，对代码的掌控不是很精准，造成许多小BUGS，有待进一步的加强。