题目描述
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 1010 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。
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----迷宫-----
import java.util.Scanner; import java.util.*; public class test1 { //输入的迷宫信息 static char[][] graph = new char[30][50]; //标记是否已经走过 static boolean[][] visited = new boolean[30][50]; //定义起点为(0,0) 终点为(29,49); //D(1,0) L(0,-1) R(0,1) U(-1,0) static int[] dir_x = {1,0,0,-1}; static int[] dir_y = {0,-1,1,0}; static String[] str = {"D","L","R","U"}; public static void main(String[] args) { Scanner scan = new Scanner(System.in); //读入数据 一行行读 转字符数组 for(int i = 0; i < 30; i++){ graph[i] = scan.nextLine().toCharArray(); } bfs(); scan.close(); } //广度优先遍历 按照字典序 先找到的就是最短的且最小的 public static void bfs() { //使用队列进行遍历 Queue<Node> que = new LinkedList<>(); que.offer(new Node(0,0,"")); //一些队列有大小的限制,因此在一个满的队列中加入一个新的选项,多出的选项就会被拒绝,offer()方法他不是对调用add()方法返回一个unchecked异常,而是得到由offer()返回的false visited[0][0] = true; while(!que.isEmpty()){ //判断que是否为空 空 TRUE //获取队首结点 Node node = que.poll();//poll方法从队列中删除第一个元素,用在空集合中不会异常抛出,只是会返回null if(node.x == 29 && node.y == 49){ System.out.println(node.path); return; } for(int i = 0; i < 4; i++){ int x = node.x + dir_x[i]; int y = node.y + dir_y[i]; if(x >= 0 && x <= 29 && y >= 0 && y <= 49 && !visited[x][y] && graph[x][y] == '0'){ visited[x][y] = true; que.offer(new Node(x,y,node.path + str[i])); } } } } } //迷宫结点,记录当前位置以及沿途路径 class Node{ int x,y; String path; public Node(int x, int y, String path){ this.x = x; this.y = y; this.path = path; } }
标签:node,int,迷宫,static,new,que From: https://www.cnblogs.com/mcpf/p/17054571.html