Minesweeper

题目描述
Minesweeper Have you ever played Minesweeper? This cute little game comes with a certain operating system whose name we can't remember. The goal of the game is to find where all the mines are located within a M x N field. The game shows a number in a square which tells you how many mines there are adjacent to that square. Each square has at most eight adjacent squares. The 4 x 4 field on the left contains two mines, each represented by a ``*'' character. If we represent the same field by the hint numbers described above, we end up with the field on the right: *... .... .*.. .... *100 2210 1*10 1110
(
扫雷艇你玩过扫雷艇?这个可爱的小游戏有一个特定的操作系统,我们不能忘了他的名字。游戏的目的是找出所有的矿山都位于一个M x N场。游戏展示了一个广场,告诉你数量有多少地雷广场毗邻。每平方最多8个相邻方块。左边的4 x 4场包含两个矿山,每个代表一个“*”字符。如果我们代表相同的字段的提示数字上面所描述的那样,我们得到右边的领域:*…....。* . . ....* 1 * 10 100 100 2210
)
输入
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 stand for the number of lines and columns of the field, respectively. Each of the next n lines contains exactly m characters, representing the field. Safe squares are denoted by ``.'' and mine squares by ``*,'' both without the quotes. The first field line where n = m = 0 represents the end of input and should not be processed.
(
输入将包含任意数量的字段。每个字段的第一行包含两个整数n和m(0 < n,m < 100),代表的行数和列的字段,分别。每一个下一个n行包含m字符,代表。安全的方块用' '。“我和方块“*”,既没有引号。第一个字段行n = m = 0代表输入和不应该处理的结束。
)
输出
For each field, print the message Field #x: on a line alone, 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 mines adjacent to that square. There must be an empty line between field outputs.
(
每个字段,打印消息字段# x:在单独一行,其中x代表字段的数量从1开始。接下来的n行应该包含字段' '。”字符取代了煤矿相邻,广场的数量。字段之间必须有一个空行输出。
)
样例输入
4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0

样例输出
Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100

 

 

#include <stdio.h>
#include <string.h>
int main()
{
    char lei[120][120];
    int ci=0,n,m;
    while(~scanf("%d%d",&n,&m)&&(n||m))
    {
        memset(lei,'0',sizeof(lei));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                char x;
                scanf(" %c",&x);
                if(x=='*')
                {
                    lei[i][j]='*';
                    for(int ii=i-1; ii<=i+1; ii++)
                        for(int jj=j-1; jj<=j+1; jj++)
                            if(lei[ii][jj]!='*')lei[ii][jj]++;
                }
            }
        printf("Field #%d:\n",++ci);
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)printf(j!=m?"%c":"%c\n",lei[i][j]);
        printf("\n");
    }
    return 0;
}