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, m100) 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;
}