hdu1241 Oil Deposits

从今天巩固一下搜索方面的题，从简单的做起吧
http://acm.hdu.edu.cn/showproblem.php?pid=1241

Problem Description
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid. 


Input
The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.


Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.


Sample Input
1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5 
****@
*@@*@
*@**@
@@@*@
@@**@
0 0 


Sample Output
0
1
2
2

题意，@为油井，@周围八个格若也为@，则他们属于同一块油田，计算油田总数。

#include <iostream>

using namespace std;
char c[105][105];
int n,m;
int f[8][2]={-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};
void dfs(int a,int b)
{
    int x,y;
    c[a][b]='*';//马上变为不是油井，防止再次搜索时重复
    for(int i=0;i<8;i++)八个方向查找
    {
        x=a+f[i][0];
        y=b+f[i][1];
        if(x>=0&&y>=0&&x<n&&y<m&&c[x][y]=='@')
        dfs(x,y);
    }
}
int main()
{while(cin>>n>>m)
 {
     if(n==0&&m==0)break;
     for(int i=0;i<n;i++)
     for(int j=0;j<m;j++)
     cin>>c[i][j];
     int sum=0;
     for(int i=0;i<n;i++)
     {
         for(int j=0;j<m;j++)
         {
             if(c[i][j]=='@')
             {
                 sum++;
                 dfs(i,j);
             }
         }
     }
     cout<<sum<<endl;
 }

    return 0;
}