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 <string.h>
int table[201][201];
void add(int x, int y, int n, int m) {
	int i, dx, dy;
	int D[8][2] = {{1,0},{0,1},{-1,0},{0,-1},
					{1,1},{1,-1},{-1,1},{-1,-1}};
	for(i = 0; i < 8; i++) {
		dx = D[i][0], dy = D[i][1];
		if(x + dx >= 0 && x + dx < n && y + dy >= 0 && y + dy < m) {
			table[x+dx][y+dy]++;
		}
	}
}
int main() {
	int n, m, Case = 0, i, j;
	char map[201][201];
	while(scanf("%d %d", &n, &m) == 2) {
		if(n == 0 && m == 0)
			break;
		if(++Case != 1)	puts("");
		for(i = 0; i < n; i++)
			scanf("%s", &map[i]);
		memset(table, 0, sizeof(table));
		for(i = 0; i < n; i++) 
			for(j = 0; j < m; j++)
				if(map[i][j] == '*') {
					add(i, j, n, m);
				}
		printf("Field #%d:\n", Case);
		for(i = 0; i < n; i++, puts("")) {
			for(j = 0; j < m; j++)
				if(map[i][j] == '*')
					printf("*");
				else
					printf("%d", table[i][j]);
		}
	}
    return 0;
}