UVA10189 Minesweeper【Ad Hoc】

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 M × N 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 4 × 4 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.

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.

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

问题链接：UVA10189 Minesweeper

问题简述：（略）

问题分析：

　　扫雷游戏是人们熟知的一个计算机游戏，通过这个程序，可以了解其局部的计算是如何实现的。

　　这是一个简单的计算问题，对于一个位置，直接计算其周围的地雷数量即可。

程序说明：

　　定义二维数组用于存储扫雷游戏的棋盘时，周围多出一圈，可以省去数组下标的越界判定。　　

　　数组around[]存储周围元素的相对下标，使得无序变有序，可以用循环来处理。

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* UVA10189 Minesweeper */#include <stdio.h>
#include <string.h>#define AN 8
struct _around {int drow;int dcol;
} around[] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}};#define N 100
char a[N+2][N+2];int main(void)
{int n, m, i, j, k, caseno=0;while(~scanf("%d%d", &n, &m) && (n || m)) {memset(a, 0, sizeof(a));for(i=1; i<=n; i++)scanf("%s", a[i]+1);if(caseno)printf("\n");printf("Field #%d:\n", ++caseno);for(i=1; i<=n; i++) {for(j=1; j<=m; j++) {int count = 0;if(a[i][j] != '*') {for(k=0; k<AN; k++)if(a[i + around[k].drow][j + around[k].dcol] == '*')count++;a[i][j] = count + '0';}}}for(i=1; i<=n; i++)printf("%s\n", a[i] + 1);}return 0;
}

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