POJ 1979 题解

Red and Black

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 31722		Accepted: 17298

Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles. 

Write a program to count the number of black tiles which he can reach by repeating the moves described above. 

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20. 

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows. 

'.' - a black tile 
'#' - a red tile 
'@' - a man on a black tile(appears exactly once in a data set) 
The end of the input is indicated by a line consisting of two zeros. 

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

Sample Output

45
59
6
13

Source

Japan 2004 Domestic

 

#include "cstdio"
#include "cstring"
#include "cmath"
#include "iostream"
#include "string"

using namespace std ;
const int maxN = 210 ;

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

char mp[ maxN ][ maxN ] ;
int N , M ;

void DFS ( const int xi , const int yi ) {
        if ( !mp[ xi ][ yi ] )return ;
        mp[ xi ][ yi ] = '%' ;
        for ( int i=0 ; i<4 ; ++i ) {
                int xx = xi + dx[ i ] ;
                int yy = yi + dy[ i ] ;
                if ( xx > 0 && xx <= M && yy > 0 && yy <= N && mp[ xx ][ yy ] == '.' )  
                        DFS( xx , yy ) ;
        }
}

int sumerize ( const int n , const int m ) {
        int ret ( 0 ) ; 
        for ( int i=1 ; i<=n ; ++i ) {
                for ( int j=1 ; j<=m ; ++j ) {
                        if ( mp[ i ][ j ] == '%' ) ++ret ; 
                }
        }
        return ret ;
}

int main ( ) {
        int start_x , start_y ; 
        while ( scanf ( "%d%d
" , &N , &M ) == 2 && N && M ) {
                for ( int i=1 ; i<=M ; ++i ) {
                        scanf ( "%s" , mp[ i ] + 1 ) ;
                }
                for ( int i=1 ; i<=M ; ++i ) {
                        for ( int j=1 ; j<=N+1 ; ++j ) {
                                if ( mp[ i ][ j ] == '@' ) {
                                        mp[ i ][ j ] = '.' ; 
                                        start_x = i ; 
                                        start_y = j ;
                                        goto Loop ; 
                                }
                        }
                }
                Loop :
                DFS ( start_x , start_y ) ;
                
                printf ( "%d
" , sumerize ( M , N ) ) ; 
                memset ( mp , 0 , sizeof ( mp ) ) ;
        } 
        
        return 0 ; 
} 

2016-10-21  16:28:48