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

百度翻译：由于最近的降雨，水聚集在农民约翰农场的各个地方，由一个n x m（1<=n<=100；1<=m<=100）的矩形表示。
每个广场都有水（“W”）或旱地（“.”）。农夫约翰想知道他田里形成了多少池塘。池塘是一组相连的正方形，
其中有水，一个正方形被认为是邻近八个邻居。

思路：循环农场，找到水后将与之相连的水全部抽出，用dfs搜索，计算要抽几次。注意水是可以斜着连的。

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int m,n;
string ct[100];
int fx[8]={1,-1,0,0,1,1,-1,-1},fy[8]={0,0,1,-1,-1,1,-1,1};
void dfs(int i,int j)
{
    ct[i][j]='.';
    for(int k=0;k<8;k++)
    {
        int x=i+fx[k];
        int y=j+fy[k];
        if(x>=0&&x<m&&y>=0&&y<n&&ct[x][y]=='W')
        {
            dfs(x,y);
        }
    }
}

int main()
{
    scanf("%d %d",&m,&n);
    for(int i=0;i<m;i++)//用string存放数据比较方便
    {
        cin>>ct[i];
    }
    int ans=0;//计数器
    for(int i=0;i<m;i++)
    {
        for(int j=0;j<n;j++)
        {
            if(ct[i][j]=='W')//找到第一个水后将与之相连的水都
            {                //抽出来
                dfs(i,j);
                ans++;
            }
        }
    }
    printf("%d",ans);
}