poj2386(简单的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

Hint

OUTPUT DETAILS:

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

题意：有一个大小为N*M的园子，雨后起了水。八连通的积水都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*

***

解题思路：从任意的W开始，不停把邻接的部分用'.'代替。1次dfs后与初始的这个W连接的所有W就都被替换成了'.'，因此知道图中不在出现'W'为止，总共进行dfs的次数就是最后的答案了，8个方向对应8种状态转移，每个格子作为dfs的参数之多被调用一次，所以复杂度为O(8*n*m)=O(n*m)。

附上代码：

 1 #include<iostream>
 2 #include<cstdio>
 3 using namespace std;
 4 char map[105][105];
 5 int n,m;
 6 int dir[8][2]={{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
 7 //现在位置为(x,y) 
 8 void dfs(int x,int y)
 9 {
10     //将现在所在位置替换为. 
11     map[x][y]='.';
12     //循环遍历移动的8个方向 
13     for(int i=0;i<8;i++)
14     {
15         int dx=x+dir[i][0];  
16         int dy=y+dir[i][1];
17         //判断(dx,dy)内是否又水 
18         if(dx>=0&&dx<n&&dy>=0&&dy<m&&map[dx][dy]=='W') dfs(dx,dy);
19     }
20     
21 }
22 int main()
23 {
24     cin>>n>>m;
25     int res=0;
26     getchar();  //吸收回车符 
27     for(int i=0;i<n;i++){
28         for(int j=0;j<m;j++){
29             scanf("%c",&map[i][j]);
30         }
31         getchar();// 将回车符读掉 
32     }
33     for(int i=0;i<n;i++)
34     {
35         for(int j=0;j<m;j++)
36         {
37             //从有水的地方开始dfs 
38             if(map[i][j]=='W')
39             {
40                 dfs(i,j);
41                 res++;
42             }
43         }
44     }
45     cout<<res<<endl;
46     return 0;
47 }

poj2386(简单的dfs）

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