[HDUOJ1312]Red And Black (经典的DFS)

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Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16129    Accepted Submission(s): 9939

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

 

 

 

 

 

 

 

一道很经典的bfs题目，比较简单，适合刚接触dfs的人。

注意:

1.在初始点进行dfs,因为初始点也是一个符合要求的点，所以别忘记了加上。

 

Code:

 

import java.util.Scanner;

public class Main {

    public static int count = 0;

    public static int maxN = 0;

    public static int maxM = 0;

    public static void main( String[] args ) {
        Scanner sc = new Scanner( System.in );
        int n, m;
        while( sc.hasNext() ) {
            count = 0;
            n = sc.nextInt();
            m = sc.nextInt();
            int si = 0;
            int sj = 0;
            if( n == 0 || m == 0 )
                return;
            else {
                maxN = n;
                maxM = m;
                char[][] maps = new char[ m ][ n ];
                for( int i = 0; i < m; i++ ) {
                    String s = sc.next();
                    for( int j = 0; j < s.length(); j++ ) {
                        char c = s.charAt( j );
                        maps[ i ][ j ] = c;
                        if( c == '@' ) {
                            si = i;
                            sj = j;
                        }
                    }
                }
                dfs( maps, si, sj );
                System.out.println( count );
            }
        }
    }

    private static void dfs( char[][] maps, int i, int j ) {
        if( !cons( i, j ) )
            return;
        if( maps[ i ][ j ] == '@' || maps[ i ][ j ] == '.' ) {
            count++;
            maps[ i ][ j ] = '#';
            dfs( maps, i + 1, j );
            dfs( maps, i - 1, j );
            dfs( maps, i, j + 1 );
            dfs( maps, i, j - 1 );
        } else {
            return;
        }
    }

    public static boolean cons( int i, int j ) {
        if( i < 0 || j < 0 || i >= maxM || j >= maxN )
            return false;
        return true;
    }

}