poj2386(dfs搜索水题）

Language:Default

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42069		Accepted: 20862

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>

#include<stdio.h>
using namespace std;
int ans=0;
int N,M;
char arr[101][101];
bool vis[101][101];
int dx[3]={-1,0,1};
int dy[3]={-1,0,1};
void dfs(int i,int j)
{
    vis[i][j]=1;
    for(int a=0;a<3;a++)
    {
        for(int b=0;b<3;b++)
        {
            if(i+dx[a]<0||i+dx[a]>=N)continue;
            if(j+dy[b]<0||j+dy[b]>=M)continue;
            if(!vis[i+dx[a]][j+dy[b]]&&arr[i+dx[a]][j+dy[b]]=='W')
            {
                dfs(i+dx[a],j+dy[b]);
            }
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    cin>>N>>M;
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<M;j++)
        {
            cin>>arr[i][j];
        }
    }
    for(int i=0;i<N;i++)
    {
        for(int j=0;j<M;j++)
        {
            if(!vis[i][j]&&arr[i][j]=='W')
            {
                dfs(i,j);ans++;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}