[HDUOJ1312]Red And Black (经典的DFS)

Hot~~招聘——亚信科技，巴卡斯（杭州），壹晨仟阳(杭州)，英雄互娱（杭州）

（包括2016级新生）除了校赛，还有什么途径可以申请加入ACM校队？

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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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).

[align=left]Sample Input[/align]

6 9

....#.

.....#

......

......

......

......

......

#@...#

.#..#.

11 9

.#.........

.#.#######.

.#.#.....#.

.#.#.###.#.

.#.#..@#.#.

.#.#####.#.

.#.......#.

.#########.

...........

11 6

..#..#..#..

..#..#..#..

..#..#..###

..#..#..#@.

..#..#..#..

..#..#..#..

7 7

..#.#..

..#.#..

###.###

...@...

###.###

..#.#..

..#.#..

0 0

[align=left]Sample Output[/align]

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;
}

}