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

CODE:

#include<bits/stdc++.h>
using namespace std;
int cnt=0;
char a[1010][1010];
int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
bool visited[1010][1010];
void dfs(int x,int y)
{
    visited[x][y]=1;
    a[x][y]='.';
    cnt++;
    for(int i=0;i<8;i++)
    {
        int tx=x+dir[i][0];
        int ty=y+dir[i][1];
        if(a[tx][ty]=='W')
        {
            dfs(tx,ty);
        }
    }
}
int main()
{
    int n,m,ans=0;
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>a[i][j];
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(a[i][j]=='W')
            {
                cnt=0;
                dfs(i,j);
                ans++;
            }
        }
    }
    cout<<ans<<endl;
}

Thanks for reading.