Lake Counting（深搜，DFS）

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

Hint

OUTPUT DETAILS: 

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

#include<iostream>
#include <deque>
#include<stack>
#include<queue>
#include<math.h>
#include<algorithm>
#include<stdlib.h>
#include<string>
#include<stdio.h>
#include<string.h>
typedef long long LL;
#define $ 3.141592654
using namespace std;
char a[100][100];
int v[100][100]={0};
int m,n;
char f(int x,int y)//判断是否越界
{
    if(x>=0&&x<m&&y>=0&&y<n)
        return a[x][y];
    return '.';
}
void DFS(int x,int y)
{
     if(v[x][y]==0)
     {
         v[x][y]=1;
         if(f(x-1,y-1)=='W')
            DFS(x-1,y-1);
         if(f(x-1,y)=='W')
            DFS(x-1,y);
         if(f(x-1,y+1)=='W')
            DFS(x-1,y+1);
         if(f(x,y-1)=='W')
            DFS(x,y-1);
         if(f(x,y+1)=='W')
            DFS(x,y+1);
         if(f(x+1,y-1)=='W')
            DFS(x+1,y-1);
         if(f(x+1,y)=='W')
            DFS(x+1,y);
         if(f(x+1,y+1)=='W')
            DFS(x+1,y+1);
     }
}
int main()
{
    int i,j,k,ans=0;
    scanf("%d%d",&m,&n);
    getchar();
    for(i=0;i<m;i++)
    {
        for(j=0;j<n;j++)
          scanf("%c",&a[i][j]);
        getchar();
    }
    for(i=0;i<m;i++)
    {
        for(j=0;j<n;j++)
        {
            if(v[i][j]==0&&a[i][j]=='W')
            {
                ans++;

                DFS(i,j);//当此次调用函数完成时，与该‘W’相邻的所有‘W’都已经被标记了；遍历的时候就不会再满足“v[i][j]==0“这个条件

            }
        }
    }
    printf("%d ",ans);
    return 0;
}