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 =  ;
 
 const int dx[  ] = {  ,  ,  , - } ;
 const int dy[  ] = {  ,  , - ,  } ;
 
 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= ; i< ; ++i ) {
                 int xx = xi + dx[ i ] ;
                 int yy = yi + dy[ i ] ;
                 if ( xx >  && xx <= M && yy >  && yy <= N && mp[ xx ][ yy ] == '.' )
                         DFS( xx , yy ) ;
         }
 }
 
 int sumerize ( const int n , const int m ) {
         int ret (  ) ;
         for ( int i= ; i<=n ; ++i ) {
                 for ( int j= ; j<=m ; ++j ) {
                         if ( mp[ i ][ j ] == '%' ) ++ret ;
                 }
         }
         return ret ;
 }
 
 int main ( ) {
         int start_x , start_y ;
         while ( scanf ( "%d%d\n" , &N , &M ) ==  && N && M ) {
                 for ( int i= ; i<=M ; ++i ) {
                         scanf ( "%s" , mp[ i ] +  ) ;
                 }
                 for ( int i= ; i<=M ; ++i ) {
                         for ( int j= ; j<=N+ ; ++j ) {
                                 if ( mp[ i ][ j ] == '@' ) {
                                         mp[ i ][ j ] = '.' ;
                                         start_x = i ;
                                         start_y = j ;
                                         goto Loop ;
                                 }
                         }
                 }
                 Loop :
                 DFS ( start_x , start_y ) ;
 
                 printf ( "%d\n" , sumerize ( M , N ) ) ;
                 memset ( mp ,  , sizeof ( mp ) ) ;
         } 
 
         return  ;
 } 

2016-10-21  16:28:48