[POJ 2386] Lake Counting(DFS)

Lake Counting

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

Java AC 代码：

import java.util.Scanner;

public class Main {
    static int n,m,ans;
    static char [][]filed = new char[105][105];
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext()) {
            n = cin.nextInt();
            m = cin.nextInt();
            String s;
            for(int i = 1;i <= n;i++) {
                s = cin.next();
                for(int j = 1;j <= m;j++) {
                    filed[i][j] = s.charAt(j - 1);
                }
            }
            Slove();
            System.out.println(ans);
        }
    }

    private static void Slove() {
        ans = 0;
        for(int i = 1;i <= n;i++) {
            for(int j = 1;j <= m;j++) {
                if(filed[i][j] == 'W') {
                    DFS(i,j);
                    ans++;
                }
            }
        }
    }

    private static void DFS(int x,int y) {
        filed[x][y] = '.';
        for(int i = -1;i <= 1;i++) {
            for(int j = -1;j <= 1;j++) {
                int nx = x + i;
                int ny = y + j;
                if(filed[nx][ny] == 'W') {
                    DFS(nx,ny);
                }
            }
        }
    }
}