POJ Lake Counting

最基本的DFS

Lake Counting

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 3

Problem 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 12W........WW..WWW.....WWW....WW...WW..........WW..........W....W......W...W.W.....WW.W.W.W.....W..W.W......W...W.......W.

 

Sample Output

3

 

Source

PKU

 

 

#include <iostream>
#include <cstring>

using namespace std;

char map[102][102];
int ans=0;

int dfs(int x,int y)
{
    if(map[x][y]=='W')
    {
        map[x][y]='.';
        dfs(x-1,y-1);  dfs(x-1,y); dfs(x-1,y+1);
        dfs(x,y-1);                      dfs(x,y+1);
        dfs(x+1,y-1); dfs(x+1,y); dfs(x+1,y+1);

        return 1;

    }
    else return 0;
}


int main()
{
    
    int m,n;
    int i,j;
    cin>>m>>n;
    ans=0;
    memset(map,'.',sizeof(map));
    for(i=1;i<=m;i++)
        for(j=1;j<=n;j++)
        {
            cin>>map[i][j];
        }


    for(i=1;i<=m;i++)
        for(j=1;j<=n;j++)
        {
            if(dfs(i,j))
                ans++;
        }

//    cout<<dfs(1,1)<<endl;

    cout<<ans;


    return 0;
}