Red and Black(简单dfs)

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12519    Accepted Submission(s): 7753


Problem 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)
 

 

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
简单dfs,无剪枝;
代码:
#include<stdio.h>
char map [ 21 ][ 21 ];
int
dir [ 4 ][ 2 ]={ 1 , 0 ,- 1 , 0 , 0 , 1 , 0 ,- 1 },step ,W ,H ;
void
dfs ( int x , int y ){
    int
tx ,ty ;
    step ++;
    for
( int i = 0 ;i < 4 ;++i ){
        tx =x +dir [i ][ 0 ];ty =y +dir [i ][ 1 ];
        if
(tx >=W ||ty >=H ||tx < 0 ||ty < 0 ||map [ty ][tx ]== '#' ||map [ty ][tx ]== '@' ) continue ;
        map [ty ][tx ]= '@' ;
        dfs (tx ,ty );
    }
}

int
main (){ int x ,y ;
    while
(scanf ( "%d%d" ,&W ,&H ),W ||H ){step = 0 ;
        for
( int i = 0 ;i <H ;++i )scanf ( "%s" ,map [i ]);
        for
( int i = 0 ;i <H ;++i ) for ( int j = 0 ;j <W ;++j ) if (map [i ][j ]== '@' )x =j ,y =i ;
        dfs (x ,y );
        printf ( "%d\n" ,step );
    }

    return
0 ;
}

你可能感兴趣的:(DFS)