POJ-2386 Lake Counting

Posted 2021-03-13 jishuren

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 56258		Accepted: 27405

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.

Source

USACO 2004 November

 

 1 #include <iostream>
 2 #include <cstdio>
 3 
 4 using namespace std;
 5 
 6 const int max_N=100+5;
 7 const int max_M=100+5;
 8 
 9 int N,M;
10 char field[max_N][max_M];
11 
12 void dfs(int x,int y)
13 {
14     field[x][y]=‘.‘;
15     for(int dx=-1;dx<=1;++dx)
16     {
17         for(int dy=-1;dy<=1;++dy)
18         {
19             int nx=x+dx;
20             int ny=y+dy;
21             if(nx>=0 && nx<N && ny>=0 && ny<M && field[nx][ny]==‘W‘)
22             {
23                 dfs(nx,ny);
24             }
25         }
26     }
27 }
28 
29 
30 void solve()
31 {
32     int res=0;
33     for(int i=0;i<N;++i)
34     {
35         for(int j=0;j<M;++j)
36         {
37             if(field[i][j]==‘W‘)
38             {
39                 dfs(i,j);
40                 ++res;
41             }
42         }
43     }
44     printf("%d
",res);
45 }
46 
47 int main()
48 {
49     scanf("%d %d",&N,&M);
50     for(int i=0;i<N;++i)
51     {
52         scanf("%s",field[i]);
53     }
54     solve();
55     return 0;
56 }