Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.
 
解题思路：
此题与二叉树遍历类似，图也有DFS和BFS遍历，这个问题用的是DFS找连通块：从每个"W“格子出发，递归遍历它周围的”W"格子。每次访问一个格子时就给它写上一个“连通分量编号”即下面代码的q数组“，这样就可以在访问之前检查它是否已经有了编号，从而避免同一个格子重复访问。
下面代码是用一个二重循环来找到当前格子的相邻8个格子，也可以用常量数组或者写8条调用语句，
程序代码：
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn=100+5;
char p[maxn][maxn];
int m,n,q[maxn][maxn];

void fld(int i,int j,int w)
{
    if(i<0||i>=m||j<0||j>=n)    return ;
    if(q[i][j]>0||p[i][j]!='W')  return;
    q[i][j]=w;
    for(int r=-1;r<=1;r++)
        for(int c=-1;c<=1;c++)
            if(c==0&&r==0)  continue;
            else  fld(r+i,j+c,w);

}
int main()
{
    while(scanf("%d%d",&m,&n)==2&&m&&n)
    {
        for(int i=0;i<m;i++)    scanf("%s",p[i]);
        memset(q,0,sizeof(q));
        int cf=0;
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
            if(q[i][j]==0&&p[i][j]=='W')    fld(i,j,++cf);
        printf("%d\n",cf);
    }
    return 0;
}