POJ 2386（DFS）

深度优先搜索属于图算法的一种，英文缩写为DFS即Depth First Search.其过程简要来说是对每一个可能的分支路径深入到不能再深入为止，而且每个节点只能访问一次.

举例说明之：下图是一个无向图，如果我们从A点发起深度优先搜索（以下的访问次序并不是唯一的，第二个点既可以是B也可以是C,D），则我们可能得到如下的一个访问过程：A->B->E（没有路了！回溯到A)->C->F->H->G->D（没有路，最终回溯到A,A也没有未访问的相邻节点，本次搜索结束）.简要说明深度优先搜索的特点：每次深度优先搜索的结果必然是图的一个连通分量.深度优先搜索可以从多点发起.如果将每个节点在深度优先搜索过程中的"结束时间"排序（具体做法是创建一个list，然后在每个节点的相邻节点都已被访问的情况下，将该节点加入list结尾，然后逆转整个链表)，则我们可以得到所谓的"拓扑排序",即topological sort.

深度优先遍历图的方法是，从图中某顶点v出发：

（1）访问顶点v；

（2）依次从v的未被访问的邻接点出发，对图进行深度优先遍历；直至图中和v有路径相通的顶点都被访问；

（3）若此时图中尚有顶点未被访问，则从一个未被访问的顶点出发，重新进行深度优先遍历，直到图中所有顶点均被访问过为止。　当然，当人们刚刚掌握深度优先搜索的时候常常用它来走迷宫.事实上我们还有别的方法，那就是广度优先搜索(BFS).

http://poj.org/problem?id=2386

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19801		Accepted: 9955

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

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int m,n;
char field[105][105];  //全局变量好
void dfs(int x,int y);
int main()
{
	while(~scanf("%d%d",&m,&n))
	{

		int i, j, t = 0;
		for(i=0;i<m;i++)
		{
			getchar(); //记得加上
			for(j=0;j<n;j++)
			{
				scanf("%c",&field[i][j]);
			}
		}
		for(i=0;i<m;i++)
		{
			for(j=0;j<n;j++)
			{
				if(field[i][j] == 'W')
				{
					dfs(i,j);
					t++;
				}
			}
		}
		printf("%d
",t);
	}
}

void dfs(int x,int y)
{
	int i, j;
	field[x][y] = '.'; //标记走过
	for(i = -1; i < 2; i++)
	{
		for(j = -1; j < 2; j++)
		{		//遍历这个店相邻的8个点
			int nx = x + i, ny = y + j;
			if(nx >=0 && nx < m && ny >= 0 && ny < n && field[nx][ny] == 'W'){
				dfs(nx,ny);
			}
		}
	}
	return;
}