Lake Counting(DFS连通图)

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.

 题目意思：查找所有由w组成的区域，八面搜索，只要周围有w就能连接在一起组成一个区域。

 解题思路：这是一道很基本的深搜例题，当成模板直接来使用吧。

 

上代码：

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #include<iostream>
 using namespace std;
 char map[][];
 int vis[][];///标记数组
 int dir[][]={{,},{-,},{,},{,-},{,},{-,-},{,-},{-,}};
 int n,m;
 void DFS(int x,int y)
 {
     int a,b,i;
     vis[x][y]=;
     for(i=;i<;i++)
     {
         a=x+dir[i][];
         b=y+dir[i][];
         if(a>=&&a<n&&b>=&&b<m&&vis[a][b]==&&map[a][b]=='W')
         {
             DFS(a,b);
         }
     }
         return ;
 }
 int main()
 {
     int count,i,j;
     memset(map,,sizeof(map));
     memset(vis,,sizeof(vis));
     scanf("%d%d",&n,&m);
     getchar();
     count=;
     for(i=;i<n;i++)
     {
         scanf("%s",map[i]);
     }
     for(i=;i<n;i++)
     {
         for(j=;j<m;j++)
         {
             if(vis[i][j]==&&map[i][j]=='W')
             {
                 count++;
                 DFS(i,j);
             }
         }
     }
     printf("%d\n",count);
     return ;
 }