L - Oil Deposits

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

题解：先用结构体把每一个pockets的位置存起来，然后遍历每一个pockets，用一个数组标记现在所有的我们已经找到的pockets，如果这个pockets没有被找到过，然后进行bfs，用bfs来寻找相邻的，如果这个已经在前边bfs找到了，那么就跳过这个pockets 的bfs，最后统计进行bfs的数量就行了。

代码：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>

using namespace std;

char maap[120][120];

struct node
{
    int x,y;
}h[10005];
int num[120][120];
int dsd[8][2]={{-1,0},{1,0},{0,1},{0,-1},{-1,1},{-1,-1},{1,1},{1,-1}};
int a,b;

void bfs(struct node s)
{
    queue<node>Q;
    Q.push(s);
    num[s.x][s.y]=1;
    while(Q.size())
    {
        s=Q.front();
        Q.pop();
        int t;
        struct node p;
        for(t=0;t<8;t++)
        {
            p=s;
            p.x+=dsd[t][0];
            p.y+=dsd[t][1];
            if(p.x>=1&&p.x<=a&&p.y>=1&&p.y<=b&&num[p.x][p.y]==0&&maap[p.x][p.y]=='@')
            {
                num[p.x][p.y]=1;
                Q.push(p);
            }
        }
    }
}

int main()
{
    int c,d,e,f,g;
    while(~scanf("%d %d",&a,&b)&&a)
    {
        getchar();
        e=-1;
        for(c=1;c<=a;c++)
        {
            for(d=1;d<=b;d++)
            {
                scanf("%c",&maap[c][d]);
                if(maap[c][d]=='@')
                {
                    h[++e].x=c;
                    h[e].y=d;
                }
            }
            getchar();
        }
        memset(num,0,sizeof(num));
        g=0;
        for(f=0;f<=e;f++)
        {
            if(num[h[f].x][h[f].y]==0)
            {
                g++;
                bfs(h[f]);
            }
        }
        printf("%d\n",g);
    }
    return 0;
}