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.

中文题目：

有一个大小为N×M的园子，雨后积起了水，被认为是连接在一起的。请求出园子里总共有多少水洼？（八连通指的是下图中相对W的*部分）

***

*W*

***

* 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.

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

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

3

分析：
深度优先搜索：
尽可能“深“的遍历一个图，在深度优先搜索中，对于最新已经发现的顶点，如果它的邻接顶点未被访问，则深度优先搜索该邻接顶点。
该题即利用深度优先搜索一个水洼即一次深度优先搜索，将搜索到的"W"变成".",遍历数组，能进行几次深度优先搜索就有几个水洼。
代码如下：

#include <iostream>
#include <stdio.h>
#define MAX_N 100
#define MAX_M 100
using namespace std;
int N,M;
char field[MAX_N][MAX_M+1];
void solve();
int main()
{
    while(scanf("%d%d",&N,&M)!=EOF){
        for(int i=0;i<N;i++){
            scanf("%s",field[i]);
    }
    solve();
    }
    return 0;
}
void dfs(int x,int y){
field[x][y]=‘.‘;
for(int dx=-1;dx<=1;dx++){
    for(int dy=-1;dy<=1;dy++){
        int nx=x+dx,ny=y+dy;
        if(nx>=0&&nx<N&&ny>=0&&ny<M&&field[nx][ny]==‘W‘){
            dfs(nx,ny);
        }
    }
}
return;
}
void solve()    {
    int res=0;
  for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            if(field[i][j]==‘W‘){
                 dfs(i,j);
                res++;
            }
        }
    }
    printf("%d\n",res);
}