HDOJ 1312 Red and Black (简单dfs)

此题为一个简单dfs模板题。

题目大意为：“@”表示人所在位置，“.”表示黑色瓷砖(表示可以移动到此位置)，"#"表示红色瓷砖(表示不能移动到此位置)

注意：此题输入为先是行数再是列数

题目链接:点击打开链接

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20216    Accepted Submission(s): 12308

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

具体代码实现如下：

import java.util.Scanner;

public class Main{
	
	public static char a[][];			//字符数组
	public static boolean color[][];	//标记数组
	public static final int b[][]={{0,1},{0,-1},{-1,0},{1,0}};
	//上下左右，四个方向
	public static int count,w,h;
	
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		while(sc.hasNext()){
			w=sc.nextInt();
			h=sc.nextInt();
			if(w==0&&h==0){
				break;
			}
			a=new char[h][w];
			color=new boolean[h][w];
			int x=-1,y=-1;
			for(int i=0;i=0&&nx=0&&ny