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\n",ans);
    return 0;
}

Sample Output

3