YTU 1099: Minesweeper
1099: Minesweeper
时间限制: 1 Sec 内存限制: 64 MB提交: 180 解决: 98
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
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.
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;
}
