[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);
                 }
             }
         }
     }
 }