Java实现第十届蓝桥杯迷宫

试题 E: 迷宫
本题总分：15 分
【问题描述】
下图给出了一个迷宫的平面图，其中标记为 1 的为障碍，标记为 0 的为可 以通行的地方。
010000 000100 001001 110000
迷宫的入口为左上角，出口为右下角，在迷宫中，只能从一个位置走到这 个它的上、下、左、右四个方向之一。 对于上面的迷宫，从入口开始，可以按DRRURRDDDR 的顺序通过迷宫， 一共 10 步。其中 D、U、L、R 分别表示向下、向上、向左、向右走。 对于下面这个更复杂的迷宫（30 行 50 列），请找出一种通过迷宫的方式， 其使用的步数最少，在步数最少的前提下，请找出字典序最小的一个作为答案。 请注意在字典序中D<L<R<U。（如果你把以下文字复制到文本文件中，请务 必检查复制的内容是否与文档中的一致。在试题目录下有一个文件 maze.txt， 内容与下面的文本相同）
01010101001011001001010110010110100100001000101010 00001000100000101010010000100000001001100110100101 01111011010010001000001101001011100011000000010000 01000000001010100011010000101000001010101011001011 00011111000000101000010010100010100000101100000000 11001000110101000010101100011010011010101011110111 00011011010101001001001010000001000101001110000000
10100000101000100110101010111110011000010000111010 00111000001010100001100010000001000101001100001001 11000110100001110010001001010101010101010001101000 00010000100100000101001010101110100010101010000101 11100100101001001000010000010101010100100100010100 00000010000000101011001111010001100000101010100011 10101010011100001000011000010110011110110100001000 10101010100001101010100101000010100000111011101001 10000000101100010000101100101101001011100000000100 10101001000000010100100001000100000100011110101001 00101001010101101001010100011010101101110000110101 11001010000100001100000010100101000001000111000010 00001000110000110101101000000100101001001000011101 10100101000101000000001110110010110101101010100001 00101000010000110101010000100010001001000100010101 10100001000110010001000010101001010101011111010010 00000100101000000110010100101001000001000000000010 11010000001001110111001001000011101001011011101000 00000110100010001000100000001000011101000000110011 10101000101000100010001111100010101001010000001000 10000010100101001010110000000100101010001011101000 00111100001000010000000110111000000001000000001011 10000001100111010111010001000110111010101101111000
【答案提交】
这是一道结果填空的题，你只需要算出结果后提交即可。本题的结果为一 个字符串，包含四种字母 D、U、L、R，在提交答案时只填写这个字符串，填 写多余的内容将无法得分。

package JavaB;

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;

public class migong {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		try {
			String s = "01010101001011001001010110010110100100001000101010"
					+ "00001000100000101010010000100000001001100110100101"
					+ "01111011010010001000001101001011100011000000010000"
					+ "01000000001010100011010000101000001010101011001011"
					+ "00011111000000101000010010100010100000101100000000"
					+ "11001000110101000010101100011010011010101011110111"
					+ "00011011010101001001001010000001000101001110000000"
					+ "10100000101000100110101010111110011000010000111010"
					+ "00111000001010100001100010000001000101001100001001"
					+ "11000110100001110010001001010101010101010001101000"
					+ "00010000100100000101001010101110100010101010000101"
					+ "11100100101001001000010000010101010100100100010100"
					+ "00000010000000101011001111010001100000101010100011"
					+ "10101010011100001000011000010110011110110100001000"
					+ "10101010100001101010100101000010100000111011101001"
					+ "10000000101100010000101100101101001011100000000100"
					+ "10101001000000010100100001000100000100011110101001"
					+ "00101001010101101001010100011010101101110000110101"
					+ "11001010000100001100000010100101000001000111000010"
					+ "00001000110000110101101000000100101001001000011101"
					+ "10100101000101000000001110110010110101101010100001"
					+ "00101000010000110101010000100010001001000100010101"
					+ "10100001000110010001000010101001010101011111010010"
					+ "00000100101000000110010100101001000001000000000010"
					+ "11010000001001110111001001000011101001011011101000"
					+ "00000110100010001000100000001000011101000000110011"
					+ "10101000101000100010001111100010101001010000001000"
					+ "10000010100101001010110000000100101010001011101000"
					+ "00111100001000010000000110111000000001000000001011"
					+ "10000001100111010111010001000110111010101101111000";
			int[][] labyrinth = new int[30][50];
			for (int i = 0; i < 30; i++) {
				for (int j = 0; j < 50; j++) {
					labyrinth[i][j] = s.charAt(50 * i + j) - '0';
				}
			}
			System.out.println(BFS(labyrinth, 30, 50));
		} catch (Exception e) {
			input.close();
		}
	}

	public static String BFS(int[][] labyrinth, int row, int column) {
		int[][] stepArr = { { -1, 0 }, { 0, 1 }, { 0, -1 }, { 1, 0 } };
		String[] direction = { "U", "R", "L","D"}; 
		int[][] visit = new int[row][column];// 标记是否已经访问过
		StringBuilder sb = new StringBuilder();
		Node node = new Node(0, 0, -1, -1, 0, null);
		Queue<Node> queue = new LinkedList<Node>();
		Stack<Node> stack = new Stack<Node>();
		queue.offer(node);
		while (!queue.isEmpty()) {
			Node head = queue.poll();
			stack.push(head); // 用于回溯路径
			visit[head.x][head.y] = 1;
			for (int i = 0; i < 4; i++) {
				int x = head.x + stepArr[i][0];
				int y = head.y + stepArr[i][1];
				String d = direction[i];
				// exit
				if (x == row - 1 && y == column - 1 && labyrinth[x][y] == 0 && visit[x][y] == 0) {
					// 打印路径
					Node top = stack.pop();
					sb.append(d);
					sb.append(top.direction);
					int preX = top.preX;
					int preY = top.preY;
					while (!stack.isEmpty()) {
						top = stack.pop();
						if (preX == top.x && preY == top.y) {
							if (top.direction != null)
								sb.append(top.direction);
							preX = top.preX;
							preY = top.preY;
						}

					}
					return sb.reverse().toString();
				}
				// bfs
				if (x >= 0 && x < row && y >= 0 && y < column && labyrinth[x][y] == 0 && visit[x][y] == 0) {
					Node newNode = new Node(x, y, head.x, head.y, head.step + 1, d);
					queue.offer(newNode);
				}
			}
		}
		return null;
	}
}

class Node {
	int x, y;
	int step;
	int preX, preY;
	String direction;

	Node(int x, int y, int preX, int preY, int step, String direction) {
		this.x = x;
		this.y = y;
		this.preX = preX;
		this.preY = preY;
		this.step = step;
		this.direction = direction;
	}

}