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Use java algorithm BFS to find the shortest path to the maze exit

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Release: 2020-11-10 15:38:08
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Use java algorithm BFS to find the shortest path to the maze exit

Establishment of queue

static Queue r = new LinkedList(); //创建队列
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Basic method of queue

r.offer(); Enter the end of the queue

r.poll(); Exit the first part of the queue

r.peek(); Contents of the first part of the queue

Code implementation :

Global variable setting

package Two;

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

public class BFS {
  static int a[][] = new int [100][100]; //输入迷宫
  static int v[][] = new int [100][100]; //走过的标记为1
  static int startx,starty;    //输入起点位置
  static int p,q;      //输入要到达的坐标位置
  static int dx[] = {0,1,0,-1};  //方向数组
  static int dy[] = {1,0,-1,0}; 
  
  static Queue<point> r = new LinkedList<point>();  //创建队列
  
  
  static class point{     //建立类坐标属性
	  int x;
	  int y;
	  int step;
	
  }
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Input the maze and starting position, target position

public static void main(String[] args) {
  Scanner in = new Scanner(System.in);
  int m = in.nextInt();
  int n = in .nextInt();
  for(int i=1;i<=m;i++)            //输入迷宫
  	for(int j=1;j<=n;j++)
  		a[i][j] = in.nextInt();
  
  startx = in.nextInt();
  starty = in.nextInt();    //输入目标和起始位置
  p = in.nextInt();
  q = in.nextInt();
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BFS algorithm start

1. Set the team leader

  //BFS
  point start = new point();   //定义一个初始类作为队首
  start.x = startx;
  start.y = starty;
  start.step = 0;
  r.offer(start);
  v[startx][starty]=1;
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2. Enter the loop body

  while(!r.isEmpty()) {        //当队列为空时跳出循环
  	
  	int x = r.peek().x;      //把队首的属性赋值
  	int y = r.peek().y;
  	int step = r.peek().step;
  	
  	
  	if(x==p && y==q) {           //到达目的地,退出循环
  		System.out.println(step);
  		break;
  	}
  	
  	for(int i=0;i<4;i++) {       //广度遍历,右下左上分别入队
  		int tx= x+dx[i];
  		int ty= y+dy[i];
  	

  		if(a[tx][ty] == 1 && v[tx][ty]==0) {   //判断是否可以入队
  			//入队
  			point temp = new point();    //建立一个临时类
  			temp.x = tx;
  			temp.y = ty;
  			temp.step = r.peek().step +1;
  			
  	
  			r.offer(temp);     //入队
  			v[tx][ty]=1;       //标记为1
  		}
  	}
  	
  	r.poll(); //拓展完了需要队首出队
  
  	
  }
  

}
}
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