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 = 210 ;

const int dx[ 10 ] = { 0 , 0 , 1 , 1 , 1 , -1 , -1 , -1 } ;
const int dy[ 10 ] = { 1 , -1 , 1 , 0 , -1 , 1 , 0 , -1 } ;

int mp[ maxN ][ maxN ] ;

void DFS ( const int xi , const int yi ) {
        if ( !mp[ xi ][ yi ] )return ;
        mp[ xi ][ yi ] = false ;
        for ( int i=0 ; i<8 ; ++i ) {
                int xx = xi + dx[ i ] ;
                int yy = yi + dy[ i ] ;
                DFS( xx , yy ) ;
        }
}

int main ( ) {
        int N , M  ;
        int _cnt = 0 ; 
        scanf ( "%d%d" , &N , &M ) ; 
        getchar ( ) ;
        for ( int i=1 ; i<=N ; ++i ) {
                for ( int j=1 ; j<=M ; ++j ) {
                        if ( getchar ( ) == 'W' ) mp[ i ][ j ] = true ; 
                }
                getchar ( ) ;
        }
                
        for ( int i=1 ; i<=N ; ++i ) {
                for ( int j=1 ; j<=M ; ++j ) {
                        if ( mp[ i ][ j ] ) {
                                _cnt ++ ;
                                DFS ( i , j ) ;
                        }
                }
        }
        cout << _cnt << endl ; 
        return 0 ; 
} 

 

2016-10-19 00:25:45

 

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