搜索（A - Lake Counting）

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

Sample Output

3

题目大意就是给出一片地，地里有‘W’和‘.’ ‘w’代表水，‘.’代表没水的地方，让找出这一片地里有几块地方没有水。

Hint

OUTPUT DETAILS: 

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

代码如下：

#include<stdio.h>int n,m;char a[101][101];void dfs(int x,int y){    if(x<0 || x>n || y<0 || y>m || a[x][y]!='W')    return;    else    {        a[x][y]='.';        dfs(x-1,y);        dfs(x-1,y+1);        dfs(x-1,y-1);        dfs(x,y-1);        dfs(x,y+1);        dfs(x+1,y+1);        dfs(x+1,y-1);        dfs(x+1,y);    }}int main(){    int i,j,count=0;    //while(scanf("%d%d",&n,&m)!=EOF)    //{        scanf("%d%d",&n,&m);        for(i=0;i<n;i++)        scanf("%s",a[i]);        for(i=0;i<n;i++)        for(j=0;j<m;j++)        if(a[i][j]=='W')        {            count++;            dfs(i,j);        }      printf("%d\n",count);   // }    return 0;