ＤＦＳ深度优先搜索之ｌａｋｅ ｃｏｕｎｔｉｎｇ

A - Lake Counting

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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

Hint

OUTPUT DETAILS: 

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

深度优先搜索DFS、

深度优先搜索，就是对于棵树而言，对于一个子树先进行遍历，一直搜索树的孩子节点，深度优先，一直到树的叶子节点，再进行回溯

对于此题，当找到一个属于‘W'时，就对该区域的八个方向就行搜索，如果搜索到水域（x1，y1），就让该水域变为’.'，并继续搜索（x1，y1）的周围节点，一直到没有就跳出

java代码

import java.util.Scanner;public class Main {static char[][] map;static int[][] dir={{0,-1},{-1,-1},{-1,0},{-1,1},{1,0},{1,1},{0,1},{1,-1}}; boolean[][] vis = new boolean[10000][10000];public static void dfs(int x,int y,int row,int col){char c = map[x][y];map[x][y]='.';int x1,y1;for(int i=0;i<8;i++){x1=x+dir[i][0];y1=y+dir[i][1];if(x1<=0||x1>row||y1<=0||y1>col){continue;}if(map[x1][y1]=='W'){//map[x1][y1]='.';    dfs(x1,y1,row,col);}else{continue;}}}public static void main(String[] args) {// TODO Auto-generated method stubScanner in = new Scanner(System.in);int n, m;while(in.hasNext()){int cnt=0;n = in.nextInt();m = in.nextInt();map = new char[n+1][m+1];for(int i = 1;i<=n;i++){String s=in.next();//System.out.println(s.charAt(0));for(int j=1;j<=m;j++){map[i][j]=s.charAt(j-1);}}for(int i=1;i<=n;i++){for(int j =1;j<=m;j++){if(map[i][j]=='W'){dfs(i,j,n,m);cnt++;}}}System.out.println(cnt);}}}