POJ2386经典DFS深搜

/***@ author StormMaybin*@ date 2016-09-27*/

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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

Hint
OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

DFS实现

package com.stormma.poj;import java.io.BufferedInputStream;import java.util.Scanner;public class Main2386{    /**     * @param args     */    private Scanner scan = null;    private int n;    private int m;    private char[][] maze;    public Main2386()    {        scan = new Scanner(new BufferedInputStream(System.in));        n = Integer.parseInt(scan.next());        m = Integer.parseInt(scan.next());        maze = new char [n][m];        for (int i = 0; i < n; i++)            maze[i] = scan.next().toCharArray();        int ans = 0;        for (int i = 0; i < n; i++)        {            for (int j = 0; j < m; j++)            {                if (maze[i][j] == 'W')                {                    dfs (i, j);                    ans++;                }            }        }        System.out.println(ans);    }    public void dfs (int x, int y)    {        maze[x][y] = '.';        for (int dx = -1; dx <= 1; dx++)        {            for (int dy = -1; dy <= 1; dy++)            {                int nx = x + dx;                int ny = y + dy;                if (0 <= nx && nx < n && 0 <= ny && ny < m && maze[nx][ny] == 'W')                    dfs (nx, ny);            }        }        return;    }    public static void main(String[] args)    {        // TODO Auto-generated method stub        new Main2386();    }}