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

题目大意 @代表油田 @相邻（水平 竖直 斜着 八个方向）的话代表同一个油田 给你m*n的区域求油田个数

分析  简单搜索 求连通块  bfs  dfs都可以写

AC代码

 1 #include <stdio.h>
 2 #include <math.h>
 3 #include <string.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <queue>
 8 #include <stack>
 9 #include <vector>
10 #include <algorithm>
11 const int maxn=100+5;
12 using namespace std;
13 typedef long long ll;
14 char a[maxn][maxn];
15 int m,n;
16 int idx[maxn][maxn];        //标记每个点所属连通块，同样可以记录是否访问过
17 void dfs(int r,int c,int id)
18 {
19     if(r<0||r>=m||c<0||c>=n)    //注意边界
20         return;
21     if(idx[r][c]>0||a[r][c]!=‘@‘)   //不是‘@’或已经访问过
22         return;
23     idx[r][c]=id;                  //标记满足条件的点所属连通块
24     for(int i=-1;i<=1;i++)
25     {
26         for(int j=-1;j<=1;j++)      //搜索八个方向 ,多种写法
27         {
28             if(i!=0||j!=0)          //跳过  0，0 是本身
29                 dfs(r+i,c+j,id);
30         }
31     }
32 }
33 int main(int argc, char const *argv[])
34 {
35     while(scanf("%d %d",&m,&n)==2 && m && n )
36     {
37         for (int i = 0; i < m; ++i)
38         {
39             scanf("%s",a[i]);
40         }
41         memset(idx,0,sizeof(idx));   //初始化为零，大于零说明已经标号，访问过
42         int cnt=0;
43         for(int i=0;i<m;i++)
44         {
45             for(int j=0;j<n;j++)
46             {
47                 if(idx[i][j]==0 && a[i][j]==‘@‘)  //枚举没被访问过且是‘@’的点
48                     dfs(i,j,++cnt);               //对该点进行深度优先搜索
49             }
50         }
51         printf("%d\n",cnt );
52     }
53     return 0;
54 }

UVA 572 dfs求连通块