hdu 1312 Red and Black

题意：记录连在一起的黑块数，以@字符为搜索的源头，只搜索一次。

一道简单的搜索题。。。

java代码：

import java.util.ArrayList;
import java.util.Scanner;

public class Main {
	/*
	 * w-width表示列
	 * h-high表示行
	 * Node1312表示结点数组
	 * dir表示方向，上下左右。
	 */
	static int w, h, sx, sy;
	static Node1312[][] map = new Node1312[20][20];
	static int dir[][] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };

	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) {
				return;
			}
			/*
			 * h表示行
			 * w表示列
			 */
			for (int i = 0; i < h; i++) {
				String str = sc.next();
				for (int j = 0; j < w; j++) {
					map[i][j] = new Node1312(i, j, str.charAt(j));
					if (map[i][j].c == '@') {
						sx = i;
						sy = j;
					}
				}
			}
			System.out.println(breadthFirstSearch());
		}
	}

	/*
	 * breadthFirstSearch-广度优先搜索
	 * x,y表示搜索源的坐标
	 */
	private static int breadthFirstSearch() {
		MyQueue1312<Node1312> mq = new MyQueue1312<Node1312>();
		map[sx][sy].c = '#';
		mq.add(map[sx][sy]);
		int k = 0;
		for (; !mq.isEmpty(); k++) {
			Node1312 no1 = mq.pop();
			for (int i = 0; i < 4; i++) {
				int a = no1.x + dir[i][0];
				int b = no1.y + dir[i][1];
				// 越界
				if (a < 0 || b < 0 || a >= h || b >= w) {
					continue;
				}
				// 遇到障碍物
				if (map[a].c == '#') {
					continue;
				}
				map[a][b].c = '#';
				mq.add(map[a][b]);
			}
		}
		return k;
	}

}

/*
 * Node表示结点
 */
class Node1312 {
	int x;
	int y;
	char c;

	public Node1312() {
	}

	public Node1312(int x, int y, char c) {
		this.x = x;
		this.y = y;
		this.c = c;
	}
}

/*
 * 加泛型的队列。
 * 用来做广搜题。
 */
class MyQueue1312<E> {
	ArrayList<E> al = new ArrayList<E>();
	int index;

	public MyQueue1312() {
		index = 0;
	}

	public void add(E no) {
		al.add(no);
		index++;
	}

	public E pop() {
		if (al.isEmpty()) {
			return null;
		}
		E nod = al.remove(0);
		index--;
		return nod;
	}

	public boolean isEmpty() {
		return al.isEmpty();
	}

}

Red and Black

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13468 Accepted Submission(s): 8341



Problem Description
There is a rectangular room, covered with square tiles.
这里有一个方形的房间，用方形瓷砖铺盖。
Each tile is colored either red or black.
每一块瓷砖只有两种颜色，红色和黑色。
A man is standing on a black tile.
一个人站在一块黑色的瓷砖上。
From a tile, he can move to one of four adjacent tiles.
对于每一块瓷砖，他能移动到附近四个方向的一块瓷砖上。
But he can't move on red tiles, he can move only on black tiles.

但是他不能移动到红色的瓷砖上，只能移动到黑色的瓷砖上。

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

写一个程序来计算他能够访问的黑快数，重复上述条件。

Input
The input consists of multiple data sets.
输入语句包含多组测试事件。
A data set starts with a line containing two positive integers W and H;
每一个测试数据的起始行，包含两个正整数W和H；
W and H are the numbers of tiles in the x- and y- directions, respectively.
分别表示方块的y行，x列。
W and H are not more than 20.

w和m不会超过20。

There are H more lines in the data set, each of which includes W characters.
每一个集合中有h行数据，每一行有w个字符。
Each character represents the color of a tile as follows.

每一个字符代表一种颜色。

'.' - a black tile
表示黑块

'#' - a red tile
表示红块

'@' - a man on a black tile(appears exactly once in a data set)

表示一个人，（一个集合中只有一个）；

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
Asia 2004, Ehime (Japan), Japan Domestic