Lake Counting
Time Limit: 1000MS         Memory Limit: 65536K
Total Submissions: 30088         Accepted: 15038
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

#include <iostream>
using namespace std;
int N,M;
char field[100+1][101];
void dfs(int x,int y)
{
    field[x][y]='.';
    for(int dx=-1;dx<=1;dx++)
    {
        for(int dy=-1;dy<=1;dy++)
        {
            int nx=dx+x,ny=dy+y;
            if(nx>=0&&nx<N&&ny>=0&&ny<M&&field[nx][ny]=='W') dfs(nx,ny);
        }
    }
    return ;
}
int main() {
    int res=0;
    while(cin>>N>>M)
    {
        for(int i=0;i<N;i++)
        {
            for(int j=0;j<M;j++)
                cin>>field[i][j];
        }

        for(int i=0;i<N;i++)
        {
            for(int j=0;j<M;j++)
            {
                if(field[i][j]=='W')
                {
                    dfs(i,j);
                    res++;
                }
            }
        }
        printf("%d\n",res);
    }
    return 0;
}