POJ-2386

Lake Counting
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 36208   Accepted: 17982

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.

 

题意:

八连通的'W'为一片水洼,问总共有多少片水洼。

 

对于每个'W',搜索其八连通区域,将所有找到的'W'都变成'.',直至所有点都变成'.'。

期间所进行dfs的次数就是水洼的数目。

 

AC代码:

 1 #include<bits/stdc++.h>//POJ提交注意更换头文件
 2 using namespace std;
 3 
 4 char mp[110][110];
 5 
 6 int n,m;
 7 
 8 void dfs(int i,int j){
 9     mp[i][j]='.';
10     for(int dx=-1;dx<=1;dx++){
11         for(int dy=-1;dy<=1;dy++){
12             int nx=i+dx,ny=j+dy;
13             if(0<=nx&&nx<n&&0<=ny&&ny<m&&mp[nx][ny]=='W')
14             dfs(nx,ny);
15         }
16     }
17     return ;
18 }
19 
20 int main(){
21     ios::sync_with_stdio(false);
22     while(cin>>n>>m){
23         int ans=0;
24         for(int i=0;i<n;i++){
25             for(int j=0;j<m;j++){
26                 cin>>mp[i][j];
27             }
28         }
29         for(int i=0;i<n;i++){
30             for(int j=0;j<m;j++){
31                 if(mp[i][j]=='W'){
32                     dfs(i,j);
33                     ans++;
34                 }
35             }
36         }
37         cout<<ans<<endl;
38     }
39     return 0;
40 } 

 

posted @ 2017-07-26 16:23  Kiven#5197  阅读(240)  评论(0编辑  收藏  举报