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：

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 char map[110][110];
 7 int vis[110][110];///标记数组
 8 int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}};
 9 int n,m;
10 void DFS(int x,int y)
11 {
12     int a,b,i;
13     vis[x][y]=1;
14     for(i=0;i<8;i++)
15     {
16         a=x+dir[i][0];
17         b=y+dir[i][1];
18         if(a>=0&&a<n&&b>=0&&b<m&&vis[a][b]==0&&map[a][b]=='W')
19         {
20             DFS(a,b);
21         }
22     }
23         return ;
24 }
25 int main()
26 {
27     int count,i,j;
28     memset(map,0,sizeof(map));
29     memset(vis,0,sizeof(vis));
30     scanf("%d%d",&n,&m);
31     getchar();
32     count=0;
33     for(i=0;i<n;i++)
34     {
35         scanf("%s",map[i]);
36     }
37     for(i=0;i<n;i++)
38     {
39         for(j=0;j<m;j++)
40         {
41             if(vis[i][j]==0&&map[i][j]=='W')
42             {
43                 count++;
44                 DFS(i,j);
45             }
46         }
47     }
48     printf("%d\n",count);
49     return 0;
50 }