深度优先搜索是从最开始的状态出发，遍历所有可以到达的状态。

因此可以对所有的状态进行操作，或列举出所有的状态。

Lake Counting

 POJ - 2386 

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.

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<algorithm>
#include<cmath>
#include<cstdlib>
using namespace std;

int m, n;
char garden[105][105];

void dfs(int x, int y)
{
	//将当前点取消标记，避免重复查找
	garden[x][y] = '.';

	//遍历周围的八个点
	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 && garden[nx][ny]=='W')
				dfs(nx, ny);
		}
	}
}

int main(void)
{
	while(~scanf("%d%d", &n,&m)){
		getchar();	//吸收两数字后的换行符
		memset(garden, 0, sizeof(garden));
	
		for (int i = 0; i < n;i++){
			for (int j = 0; j < m;j++)
				scanf("%c", &garden[i][j]);
			getchar();	//吸收每次输入一行后的换行符
		}

		int sum = 0;
		for (int i = 0; i < n; i++){
			for (int j = 0; j < m; j++){
				if (garden[i][j] == 'W'){
					dfs(i, j);
					sum++;
				}
			}
		}
	
		cout << sum << endl;
	}

	return 0;
}