深度优先搜索(dfs)的用法——lake counting

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.

问题描述：

有一个大小为N*M的园子，雨后积了很多水。八连通的积水被认为是在一起的。请求出园子里共有多少个水洼？(八连通是指下图中相对+的*部分)

+++

+*+

+++

解题分析：

    从任意的’+’开始，不停地把邻接的部分用’*’代替，一次dfs(深度优先遍历)遍历后，与初始的这个+所连接的所有+都会被替换成*，因此直到图中没有+为止，总共进行dfs的次数即为积水的次数。

#include
#include
#include
using namespace std;
char a[1001][1001];
int n,m;
void DFS(int x,int y)
{
    int i,j,b,c;
    for(i=-1;i<=1;i++){
        for(j=-1;j<=1;j++){
            b=i+x;//代表了八个方向  (i-1,j-1)  (i-1,j)  (i-1,j+1) (i,j-1)
            c=y+j;//		  (i,j+1)   (i+1,j-1)  (i+1,j)  (i+1,j+1)
            if(b>=0&&b=0&&y