题目描述：

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.

由于近期的降雨，雨水汇集在农民约翰的田地不同的地方。我们用一个NxM(1<=N<=100;1<=M<=100)网格图表示。每个网格中有水('W') 或是旱地('.')。一个网格与其周围的八个网格相连，而一组相连的网格视为一个水坑。约翰想弄清楚他的田地已经形成了多少水坑。给出约翰田地的示意图，确定当中有多少水坑。

 输入：

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.

第1行：两个空格隔开的整数：N 和 M 第2行到第N+1行：每行M个字符，每个字符是'W'或'.'，它们表示网格图中的一排。字符之间没有空格。

 输出：

Line 1: The number of ponds in Farmer John's field.

一行：水坑的数量

 样例输入：

10 12

W........WW.

.WWW.....WWW

....WW...WW.

.........WW.

.........W..

..W......W..

.W.W.....WW.

W.W.W.....W.

.W.W......W.

..W.......W.  

程序代码：  
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int next1[9]={0,0,0,-1,-1,-1,1,1,1};
int next2[9]={0,1,-1,0,1,-1,0,1,-1};
int n,m,ans;//扩展四面八方 
char s[105][105];
void dfs(int x,int y)
{
	int tx,ty;
	s[x][y]='.';
	for(int i=1;i<=8;i++)
	{
		tx=x+next1[i];
		ty=y+next2[i];
		if(tx<1||tx>n||ty<1||ty>m||s[tx][ty]=='.')
			continue;
		s[tx][ty]='.';
		dfs(tx,ty);
	}
}
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			cin>>s[i][j];	
	ans=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			if(s[i][j]=='W')
			{
				dfs(i,j);
				ans++;
			}
		}
	}
	printf("%d\n",ans);
	return 0;
}