UVa 10189 Minesweeper (模拟)

10189 - Minesweeper

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1130

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

有一个利用数字大小来简化代码的技巧，详见代码。

完整代码：

#include
#include

int a[120][120];

int main(void)
{
	int n, m, count = 1, i, j,ii,jj;
	while (scanf("%d%d", &n, &m),n)
	{
	    getchar();
		if (count > 1) putchar('\n');
		char ch;
		memset(a,0,sizeof(a));
		for (i = 1; i <= n; i++)
		{
			for (j = 1; j <= m; j++)
			{
				scanf("%c", &ch);
				if (ch == '*')
				{
					a[i][j] = 11;///小技巧
					for (ii = i - 1; ii <= i + 1; ii++)
						for (jj = j - 1; jj <= j + 1; jj++)
							a[ii][jj]++;
				}
			}
			getchar();
		}
		printf("Field #%d:\n", count);
		for (i = 1; i <= n; i++)
		{
			for (j = 1; j <= m; j++)
			{
				if (a[i][j] > 10) putchar('*');
				else printf("%d", a[i][j]);
			}
			putchar('\n');
		}
		count++;
	}
	return 0;
}