2019蓝桥杯 迷宫 Java 代码注释

avatar
avatar
云惠网小编
2845
文章
1
评论
2021年4月6日23:17:41 评论 27 次浏览 5343字阅读17分48秒
摘要

5.迷宫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。010000000100001001110000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

5.迷宫

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

DDDDRRURRRRRRDRRRRDDDLDDRDDDDDDDDDDDDRDDRRRURRUURRDDDDRDRRRRRRDRRURRDDDRRRRUURUUUUUUULULLUUUURRRRUULLLUUUULLUUULUURRURRURURRRDDRRRRRDDRRDDLLLDDRRDDRDDLDDDLLDDLLLDLDDDLDDRRRRRRRRRDDDDDDRR

左:

image-20210405110334798

思路

答案

  • 编写方法将长的要死的字符串自动转换为二维int数组

  • BFS

  • 使用队列计算不同路径的长度

  • 最后走最短路径并拼接返回的字符串

image-20210405110744652

package lanqiao;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.function.IntPredicate;
public class Main  {
static int n = 30,m = 50;
//迷宫数组
static int[][] maze = new int[n][m];
//路径字符串
static String path = "";
static int[][] dir = {{1,0},{0,-1},{0,1},{-1,0}};
//距离数组
static int[][] dis = new int[n][m];
//动作代表标识 可能有的小伙伴会问为什么不是URLD
//这是因为一开始我们是从终点往起点走 所以是反过来的
static char[] c = {'D','L','R','U'};
public static void main(String[] args) {
maze = convert(
"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",n,m);
solution (maze);
System.out.println(path);
prinf(dis);
}
public static void solution(int [][] way) {
Queue<Integer> queue = new LinkedList<>();
//n * m - 1是终点的坐标计算出的唯一标识
//以此类推 起点(第一行第一列)的唯一标识1*1-1为0
queue.add(n * m - 1);
while(!queue.isEmpty()) {
Integer poll = queue.poll();
for(int i = 0; i < 4; i++) {
//获取队列中元素对应的行和列
int xx = poll / m;
int yy = poll % m;
//上/下/左/右动一步
xx = xx + dir[i][0];
yy = yy + dir[i][1];
//检测能不能走 能走就加入队列 dis[][]不为0代表已经走过了
if(xx < 0 || yy < 0 || xx >= n || yy >= m || maze[xx][yy] == 1 || dis[xx][yy] != 0 ) {
continue;
}
queue.add(xx * m + yy);
dis[xx][yy]= dis[poll / m][poll % m] + 1;
if(xx == 0 && yy == 0) {
//该条路径反向找到起点,结束
break;
}
}
}
//若无这一步 最后将进入死循环
dis[n-1][m-1] = 0;
//从起点开始往终点 走最短路径获得行进字符串
int x = 0, y = 0;
while(x != n-1 || y != m-1) {
for(int i = 0; i < 4; i++) {
//与上类似
int xx = x + dir[i][0];
int yy = y + dir[i][1];
//不能走则跳过 此处不能加上dis[xx][yy] == 0条件 因为最后会无法走到终点
if(xx == -1 || yy == -1 || xx == n || yy == m || maze[xx][yy] == 1) {
continue;
}
//这里不是判断dis[xx][yy] = dis[x][y] + 1的原因是我们使用的策略是距离终点越远dis越大
if(dis[x][y] == dis[xx][yy] + 1) {
//匹配成功 移动
x = xx;
y = yy;
//记录路径
path += c[i];
//已找到对应路径
break;
}
}
}
}
public static int[][] convert(String args,int x,int y){
int[][] res = new int[x][y];
for(int i = 0; i < x; i++) {
for(int j = 0; j < y; j++) {
res[i][j] = args.charAt(j + i * y) - 48;
}
}
return res;
}
public static void prinf(int [][] res) {
for(int[] a : res) {
for(int b : a) {
System.out.printf("%3d ",b);
}
System.out.println();
}
System.out.println();
}
}

右:

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

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

DIS数组效果图

本文转自 https://blog.csdn.net/Rush6666/article/details/115439525

腾讯云618
avatar
2w 字长文爆肝 JVM 经典面试题!太顶了! java

2w 字长文爆肝 JVM 经典面试题!太顶了!

如果你是中高级程序员,那我相信你一定被面试官问过JVM。下次再被问到JVM,你直接把老周的这篇文章丢给他吧!话不多说,让我们直接进入主题吧。JVM内存结构,常见异常,调优参数,调优...
JAVA初窥-DAY08 java

JAVA初窥-DAY08

JAVA初窥-DAY08面向过程与面向对象实例化及调用方法和成员变量面向过程与面向对象面向过程:注重的是某件事情过程中的每一个步骤的实现。面向对象:把面向过程中的每一个步骤交给一个...
腾讯云618
匿名

发表评论

匿名网友 填写信息

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen: