pc110102 Minesweeper(扫雷)

这个题我都快吐血了!!!老是格式错误(Presentation Error)最后终于知道了原来是输出案例时 要空一行最后一个不要换行;由于是任意案例以0 0结束,故0 0的上一行不能有换行。AC的好纠结呀!!!!

题意:给你n*m的矩阵,‘ * ’代表雷,‘ . ’代表空白处;输出空白去有多少雷;

 

 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
 

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$ \le$100) which stand for the number of lines and columns of the field, respectively. Each of the next n lines contains exactly mcharacters, representing the field.

Safe squares are denoted by ``.'' and mine squares by ``*,'' both without the quotes. The first field line where n = m = 0represents the end of input and should not be processed.

Output

 

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.

Sample Input  

 

 

4 4
*...
....
.*..
....
3 5
**...
.....
.*...
0 0

 

Sample Output

 

 

Field #1:
*100
2210
1*10
1110

Field #2:
**100
33200
1*100
 
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#define N 105
using namespace std;
char map[N][N];
int n,m,test=0;
int b=false;

int find(int i,int j)
{
int s=0;
s=(map[i-1][j]=='*')+(map[i+1][j]=='*')+(map[i-1][j-1]=='*')+(map[i+1][j-1]=='*');
s+=(map[i][j-1]=='*')+(map[i][j+1]=='*')+(map[i-1][j+1]=='*')+(map[i+1][j+1]=='*');
return s;
}


void work()
{
int i,j;
if(b) cout<<endl; //想吐血的处理
printf("Field #%d:\n",++test);
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(map[i][j]=='.')
cout<<find(i,j);
else cout<<'*';
}
cout<<endl;
}
b=true;
}

int main()
{
while(cin>>n>>m&&(n+m))
{
int i,j;
memset(map,'.',sizeof(map));

for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
cin>>map[i][j];
work();

}
return 0;
}

posted @ 2011-09-27 16:13  mtry  阅读(291)  评论(0)    收藏  举报