Laking 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

Hint

OUTPUT DETAILS: 

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

 

思路:

　　我在做这道题目的时候遇到的最大的困难是题意了，我读了很久的描述也没有弄明白什么是水洼，看了下面的提示

，无法从样例中找到所说的三个水洼。后来终于搞明白水洼必须是连续的‘W'，也就是说一个水洼其实就是一片连起来的’

W'，弄清了题意就可以使用dfs进行求解了，从任意一个'w'开始调用dfs()会把一片连续的'w'替换掉，多次调用，直到没有

'W'为止.这样调用dfs()的次数就是答案

易错：

　　其中的'W'是大写，要看清楚

import java.util.Scanner;
import java.util.Stack;

public class Main {
    static Scanner scanner = new Scanner(System.in);
    static int n;
    static int m;
    static char [][] a;
    public static void main(String[] args) {
        n = scanner.nextInt();
        m = scanner.nextInt();
        a = new char[n][m];
        for(int i = 0;i<n;i++){
            String string = scanner.next();
            a[i] = string.toCharArray();
        }

        int ans = 0;
        for(int i = 0;i<n;i++){
            for(int j = 0;j<m;j++){
                if(a[i][j]=='W'){
                    dfs(i,j);
                    ans++;
                }
            }
        }
        System.out.println(ans);
    }

    static void dfs(int x,int y){
        if(a[x][y] == 'W'){
            a[x][y] = '.';
        }

        for(int dx = -1;dx<=1;dx++){
            for(int dy = -1;dy<=1;dy++){
                int i = x + dx;
                int j = y + dy;
                if(i>=0&&i<n&&j>=0&&j<m&&a[i][j]=='W'){
                    dfs(i,j);
                }
            }
        }

    }
}