POJ 2386 题解

Lake Counting

描述

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.

输入

* 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.

输出

* Line 1: The number of ponds in Farmer John's field.

样例输入

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

样例输出

3

提示

OUTPUT DETAILS:

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

来源

USACO 2004 November

 #include "bits/stdc++.h"
 
 using namespace std ;
 const int maxN =  ;
 
 const int dx[  ] = {  ,  ,  ,  ,  , - , - , - } ;
 const int dy[  ] = {  , - ,  ,  , - ,  ,  , - } ;
 
 int mp[ maxN ][ maxN ] ;
 
 void DFS ( const int xi , const int yi ) {
         if ( !mp[ xi ][ yi ] )return ;
         mp[ xi ][ yi ] = false ;
         for ( int i= ; i< ; ++i ) {
                 int xx = xi + dx[ i ] ;
                 int yy = yi + dy[ i ] ;
                 DFS( xx , yy ) ;
         }
 }
 
 int main ( ) {
         int N , M  ;
         int _cnt =  ;
         scanf ( "%d%d" , &N , &M ) ;
         getchar ( ) ;
         for ( int i= ; i<=N ; ++i ) {
                 for ( int j= ; j<=M ; ++j ) {
                         if ( getchar ( ) == 'W' ) mp[ i ][ j ] = true ;
                 }
                 getchar ( ) ;
         }
 
         for ( int i= ; i<=N ; ++i ) {
                 for ( int j= ; j<=M ; ++j ) {
                         if ( mp[ i ][ j ] ) {
                                 _cnt ++ ;
                                 DFS ( i , j ) ;
                         }
                 }
         }
         cout << _cnt << endl ;
         return  ;
 } 

2016-10-19 00:25:45

(完)