Lake Counting --POJ 2386

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>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e2+10;
char a[maxn][maxn];
int cnt=0;
int n,m;

void DFS(int i,int j){
    for( int p=-1; p<=1; p++ ){
        for( int q=-1; q<=1; q++ ){
            int x=i+p;
            int y=j+q;
            if(x>=0&&x<=n-1&&y>=0&&y<=m-1&&a[x][y]=='W'){
                a[x][y]='.';
                DFS(x,y);
            }
        }
    }
}

int main(){
    while(~scanf("%d%d",&n,&m)&&n&&m){
        memset(a,'0',sizeof(a));
        cnt=0;
        for( int i=0; i<n; i++ ){
            scanf("%s",a[i]);
        }
        for( int i=0; i<n; i++){
            for( int j=0; j<m; j++ ){
                if(a[i][j]=='W'){
                    cnt++;
                    DFS(i,j);
                }
            }
        }
        printf("%d
",cnt);
    }
    return 0;
}