DFS----Lake Counting (poj 2386)

Lake Counting(POJ No.2386)

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

 

 1 #include <iostream>
 2 using namespace std;
 3 int N,M;
 4 //int res=0;
 5 const int MAX_N=1000;
 6 const int MAX_M=1000;
 7 char field[MAX_N][MAX_M];
 8 void dfs(int x,int y)
 9 {
10     field[x][y]='.';
11     for(int dx=-1;dx<=1;dx++)
12     {
13         for(int dy=-1;dy<=1;dy++)
14         {
15             int nx=dx+x,ny=dy+y;
16             if(nx>=0&&nx<N&&ny>=0&&ny<M&&field[nx][ny]=='W')
17                 dfs(nx,ny);
18         }
19     }
20     return;
21 }
22 void solve()
23 {
24     int res=0;
25     for(int i=0;i<N;i++) {
26         for(int j=0;j<M;j++){
27             if(field[i][j]=='W') {
28                 dfs(i, j);
29                 res++;
30             }
31         }
32     }
33     cout<<res<<endl;
34 }
35 int main() {
36     cin>>N>>M;
37     for(int x=0;x<N;x++)
38     {
39         for(int y=0;y<M;y++)
40         {
41             cin>>field[x][y];
42         }
43        // printf("\n");
44     }
45     solve();
46     //cout<<res<<endl;
47     return 0;
48 }