Lake Counting（dfs）

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.

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
char a[100][100];
int m=0,n=0;
int ad(int x,int y)
{
    a[x][y]='.';
    for(int i=-1; i<2; i++)
        for(int j=-1; j<2; j++)
            if(a[x+i][y+j]=='W'&&x+i>=0&&x+i<n&&y+j>=0&&y+j<m)
                ad(x+i,y+j);
}

int main()
{
    int ans=0;
    scanf("%d%d",&n,&m);
    getchar();
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<m; j++)
            scanf("%c",&a[i][j]);
            getchar();
    }

    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            if(a[i][j]=='W')
            {
                ad(i,j);
                ans++;
            }
    printf("%d
",ans);
    return 0;
}

Sample Output
3