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 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

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

Source

Mid-Central USA 1997

 題目大意：給出一個R*C的圖，其中@代表油田，*則不是，問你其中的油田數目是多少，其中上下左右等八個方向相鄰的油田均是一塊，也就是聯通塊

思路：很簡單的一道dfs的題目了，就是從每一個點查詢他的八個方向，如果八個方向有油田，就把這個油田變成*，這樣下一次就不會再次計算他，如果沒有，就不需要做什麼改變，這樣在下面統計結果的時候只需要判斷每一個點是否是有油田就行了，

程式碼:

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#define maxn 1001
using namespace std;

int R,C;
int cnt;
char mp[maxn][maxn];

void dfs(int x,int y)
{
    if(x<0||x>=R||y<0||y>=C||mp[x][y]!='@')
        return ;

    mp[x][y]='*';
    dfs(x+1,y);
    dfs(x,y+1);
    dfs(x-1,y);
    dfs(x,y-1);
    dfs(x+1,y+1);
    dfs(x-1,y-1);
    dfs(x+1,y-1);
    dfs(x-1,y+1);
}

int main()
{
    while(scanf("%d%d",&R,&C)==2&&R)
    {
        cnt=0;
        for(int i=0;i<R;i++)
            scanf("%s",mp[i]);

        for(int i=0;i<R;i++)
            for(int j=0;j<C;j++)
                if(mp[i][j]=='@')
                    dfs(i,j),cnt++;
        printf("%d\n",cnt);
    }
    return 0;
}