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广度优先搜索不返回最短路径

[英]Breadth first search is not returning the shortest path

I am trying to use the breadth first search algorithm using Java. 我正在尝试使用Java的广度优先搜索算法。 Considering the 10x10 grid I am trying to find the last cell 9x9 (The grid starts from 0,0 ). 考虑到10x10网格,我试图找到最后一个单元9x9(网格从0,0开始)。 By the time when it reaches the 9x9 it has traversed all the cells in the grid. 到达9x9时,它已遍历网格中的所有单元格。 I heard like the BFS will give me the shortest path. 我听说BFS会给我最短的路径。 But actually it has given me the longest path. 但实际上它给了我最长的路径。

  1. Could you please tell me if this is the expected behaviour ? 你能否告诉我这是否是预期的行为?
  2. If this is how the BFS will work then what is the best way to get the shortest route to the 9x9 the cell ? 如果这是BFS的工作方式,那么获得9x9单元最短路径的最佳方法是什么?

Please advice. 请指教。

Edit-- I have used this logic and completed my game. 编辑 - 我已经使用了这个逻辑并完成了我的游戏。 If you want to refer please check https://play.google.com/store/apps/details?id=com.game.puzzle.game.ballmania.android 如果您想参考,请访问https://play.google.com/store/apps/details?id=com.game.puzzle.game.ballmania.android

Code

package com.example.game.bfs.alogrithm;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

public class BFS {

static class Cell {
    private int x;
    private int y;
    private String value;
    private boolean visitStatus;

    public Cell(int x, int y, String value,boolean visitStatus) {
        this.x = x;
        this.y = y;
        this.value = value; 
        this.visitStatus=visitStatus;
    }
}

private Cell[][] board;

private List<Cell> visited = new ArrayList<Cell>();

private boolean testDone;

public void setBoard(Cell[][] board) {
    this.board = board;
} 

public Cell getAdjacentUnvisitedCell(Cell cell)
{  
   int moves[][] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };
   for (int n = 0; n < 4 /* moves.length */; ++n) {
       int ti = cell.x + moves[n][0];
       int tj = cell.y + moves[n][1];
      // System.out.println("ti,tj" + ti +"," + tj );  

       if (ti >= 0 && ti < board.length && tj >= 0 && tj < board[0].length) { 

          // System.out.println("getAdjacentUnvisitedCell : " + "[" + board[ti][tj].x + "," + board[ti][tj].y + "]" ); 
          // System.out.println("getAdjacentUnvisitedCell : board[ti][tj].visitStatus " + board[ti][tj].visitStatus ); 

           if(!board[ti][tj].visitStatus) {  
              return board[ti][tj];
           }
       }
   }  
   return null;  
} 

public void BFSearch(Cell start, Cell end) {  
   // BFS uses Queue data structure 
   Queue<Cell> q = new LinkedList<Cell>(); 
   q.add(start);
   visited.add(start);
   board[start.x][start.y].visitStatus = true;

   //printNode(start);

   while( !q.isEmpty() )
   { 
      Cell c; 
      c = q.peek(); 
      Cell unVisitedadjCell = getAdjacentUnvisitedCell(c); 

      if(!testDone){
          testDone=true;  
      } 

      if ( unVisitedadjCell != null )
      {  visited.add(unVisitedadjCell); 
         board[unVisitedadjCell.x][unVisitedadjCell.y].visitStatus = true;

         printNode(unVisitedadjCell,c); 
         q.add(unVisitedadjCell);
      }
      else
      {
         q.remove();
      }
   }

   visited.clear();     //Clear visited property of nodes
}


private void printNode(Cell c,Cell node) {
    System.out.println("For Node " + node.x +"," + node.y + ",   " + "Just Visited : " + "[" + c.x + "," + c.y + "]" );  
} 

public static void main(String[] args) {
    Cell[][] cells = new Cell[10][10];
    for (int i = 0; i < 10; i++) {
        for (int j = 0; j < 10; j++) {
            cells[i][j] = new Cell(i, j, "defaultvalue",false);
        }
    } 

    BFS board = new BFS();
    board.setBoard(cells);

    board.BFSearch(cells[0][0], cells[1][4]);
}


}

}

Log : 记录:

For Node 0,0,   Just Visited : [1,0]
For Node 0,0,   Just Visited : [0,1]
For Node 1,0,   Just Visited : [2,0]
For Node 1,0,   Just Visited : [1,1]
For Node 0,1,   Just Visited : [0,2]
For Node 2,0,   Just Visited : [3,0]
For Node 2,0,   Just Visited : [2,1]
For Node 1,1,   Just Visited : [1,2]
For Node 0,2,   Just Visited : [0,3]
For Node 3,0,   Just Visited : [4,0]
For Node 3,0,   Just Visited : [3,1]
For Node 2,1,   Just Visited : [2,2]
For Node 1,2,   Just Visited : [1,3]
For Node 0,3,   Just Visited : [0,4]
For Node 4,0,   Just Visited : [5,0]
For Node 4,0,   Just Visited : [4,1]
For Node 3,1,   Just Visited : [3,2]
For Node 2,2,   Just Visited : [2,3]
For Node 1,3,   Just Visited : [1,4]
For Node 0,4,   Just Visited : [0,5]
For Node 5,0,   Just Visited : [6,0]
For Node 5,0,   Just Visited : [5,1]
For Node 4,1,   Just Visited : [4,2]
For Node 3,2,   Just Visited : [3,3]
For Node 2,3,   Just Visited : [2,4]
For Node 1,4,   Just Visited : [1,5]
For Node 0,5,   Just Visited : [0,6]
For Node 6,0,   Just Visited : [7,0]
For Node 6,0,   Just Visited : [6,1]
For Node 5,1,   Just Visited : [5,2]
For Node 4,2,   Just Visited : [4,3]
For Node 3,3,   Just Visited : [3,4]
For Node 2,4,   Just Visited : [2,5]
For Node 1,5,   Just Visited : [1,6]
For Node 0,6,   Just Visited : [0,7]
For Node 7,0,   Just Visited : [8,0]
For Node 7,0,   Just Visited : [7,1]
For Node 6,1,   Just Visited : [6,2]
For Node 5,2,   Just Visited : [5,3]
For Node 4,3,   Just Visited : [4,4]
For Node 3,4,   Just Visited : [3,5]
For Node 2,5,   Just Visited : [2,6]
For Node 1,6,   Just Visited : [1,7]
For Node 0,7,   Just Visited : [0,8]
For Node 8,0,   Just Visited : [9,0]
For Node 8,0,   Just Visited : [8,1]
For Node 7,1,   Just Visited : [7,2]
For Node 6,2,   Just Visited : [6,3]
For Node 5,3,   Just Visited : [5,4]
For Node 4,4,   Just Visited : [4,5]
For Node 3,5,   Just Visited : [3,6]
For Node 2,6,   Just Visited : [2,7]
For Node 1,7,   Just Visited : [1,8]
For Node 0,8,   Just Visited : [0,9]
For Node 9,0,   Just Visited : [9,1]
For Node 8,1,   Just Visited : [8,2]
For Node 7,2,   Just Visited : [7,3]
For Node 6,3,   Just Visited : [6,4]
For Node 5,4,   Just Visited : [5,5]
For Node 4,5,   Just Visited : [4,6]
For Node 3,6,   Just Visited : [3,7]
For Node 2,7,   Just Visited : [2,8]
For Node 1,8,   Just Visited : [1,9]
For Node 9,1,   Just Visited : [9,2]
For Node 8,2,   Just Visited : [8,3]
For Node 7,3,   Just Visited : [7,4]
For Node 6,4,   Just Visited : [6,5]
For Node 5,5,   Just Visited : [5,6]
For Node 4,6,   Just Visited : [4,7]
For Node 3,7,   Just Visited : [3,8]
For Node 2,8,   Just Visited : [2,9]
For Node 9,2,   Just Visited : [9,3]
For Node 8,3,   Just Visited : [8,4]
For Node 7,4,   Just Visited : [7,5]
For Node 6,5,   Just Visited : [6,6]
For Node 5,6,   Just Visited : [5,7]
For Node 4,7,   Just Visited : [4,8]
For Node 3,8,   Just Visited : [3,9]
For Node 9,3,   Just Visited : [9,4]
For Node 8,4,   Just Visited : [8,5]
For Node 7,5,   Just Visited : [7,6]
For Node 6,6,   Just Visited : [6,7]
For Node 5,7,   Just Visited : [5,8]
For Node 4,8,   Just Visited : [4,9]
For Node 9,4,   Just Visited : [9,5]
For Node 8,5,   Just Visited : [8,6]
For Node 7,6,   Just Visited : [7,7]
For Node 6,7,   Just Visited : [6,8]
For Node 5,8,   Just Visited : [5,9]
For Node 9,5,   Just Visited : [9,6]
For Node 8,6,   Just Visited : [8,7]
For Node 7,7,   Just Visited : [7,8]
For Node 6,8,   Just Visited : [6,9]
For Node 9,6,   Just Visited : [9,7]
For Node 8,7,   Just Visited : [8,8]
For Node 7,8,   Just Visited : [7,9]
For Node 9,7,   Just Visited : [9,8]
For Node 8,8,   Just Visited : [8,9]
For Node 9,8,   Just Visited : [9,9]

The pattern in which the cells are visited. 访问单元格的模式。

在此输入图像描述

Trace back through the log, from the end to the beginning. 从末尾到开头追溯日志。 You'll see that it in fact has found the shortest path - along the edge of the grid. 你会发现它实际上找到了最短的路径 - 沿着网格的边缘。 Unfortunately in grids, if you don't allow going through diagonals (in which case BFS goes out of the window as the diagonal should have a different weight), all the paths which have only operations "to the right" and "down" are the shortest. 不幸的是,在网格中,如果你不允许通过对角线(在这种情况下BFS离开窗口,因为对角线应该具有不同的权重),所有只有“向右”和“向下”操作的路径都是最短的。

You can see it by simple logic - to go from 0 to 9 you have to make 9 moves. 你可以通过简单的逻辑看到它 - 从0到9,你必须进行9次移动。 You have 2 coordinates, you go from (0, 0) to (9, 9) , you can only change one coordinate by 1 in 1 operation, so the shortest path has 9+9=18 steps. 你有2个坐标,你从(0, 0)(9, 9) ,你只能在一个操作中改变一个坐标,所以最短的路径有9+9=18步。 Trace back and see that this path has 18 steps. 追溯并看到此路径有18个步骤。 Similarly any path from beginning to the end which only has operations to the right and down will have 18 steps, so any path like that will be the shortest. 同样,从开始只具有操作结束的任何路径to the rightdown将有18个步骤,因此,任何路径一样,这将是最短的。 What determines the path itself is simply the order in which you put the adjacent coordinates into the queue. 决定路径本身的只是将相邻坐标放入队列的顺序。 Try doing it in random order. 尝试以随机顺序执行此操作。

edit: Here's how to count the number of the shortest paths. 编辑:这是如何计算最短路径的数量。 We've previously noticed that there are 18 operations; 我们之前已经注意到有18个操作; 9 of which are to the right and 9 are down . 其中9个to the right ,9个在down Order of these operations doesn't matter because in the end you've added (9, 9) to the initial (0, 0) , so you actually arrive at the end. 这些操作的顺序无关紧要,因为最后你已经将(9, 9)添加到初始(0, 0) ,所以你实际上到达了最后。 How do we count them? 我们如何算他们? Let us assign each operation an identifier as such: a_1, a_2, ... a_18 . 让我们为每个操作分配一个标识符: a_1, a_2, ... a_18 We are now going to choose 9 of these operations to be down . 我们现在要选择这些操作的9要down So we choose the first spot for a down operation, which we can do in 18 ways (since there are 18 operations to choose), then the second ( 17 ways) and so on until we're out of down operations. 所以我们选择第一个down操作点,我们可以用18种方式做(因为有18种操作可供选择),然后是第二种( 17种方式),依此类推,直到我们完成down操作。 We could have done that in 18*17*...*10 ways. 我们本可以用18*17*...*10种方式做到这一点。 Now we choose spots for right operations. 现在我们为right操作选择点。 We can do so (by aology) in 9*8*...*1 ways. 我们可以通过9*8*...*1种方式(通过神学)这样做。 But now we don't really make a distinction between each of the down instructions, do we? 但是现在我们并没有真正区分每个down指令,对吗? We could have chosen the first down operation in 9 ways, the second in 8 and so on. 我们也可以选择第一down操作9的方式,第二次在8等。 Similarly we could have chosen right operations. 同样,我们可以选择right操作。 Finally we deduct that there are (18*17*...*1)/((9*8*...*1)*(9*8*...*1)) = 48 620 ways of doing so (we divide by the meaningless for us distinction of operations). 最后我们推断出有(18*17*...*1)/((9*8*...*1)*(9*8*...*1)) = 48 620这样做的方式(我们除以操作的区别是无意义的)。 It's also the number of ways you can choose 9 out of 18 spots. 这也是您可以选择18 9 18点的方式。

If my explanation is too messy for you I can recommend taking a look at Introductory combinatorics by Richard A. Brualdi . 如果我的解释对你来说太乱了,我可以推荐看看Richard A. Brualdi Introductory combinatorics It's a really cool book on interesting things regarding some fields of discrete maths. 关于某些离散数学领域的有趣事情,这是一本非常酷的书。 It's quite easy to read. 这很容易阅读。

Based on what i understood of other answer (this of lared ) here what you code could look like : 基于我对其他答案(这是lared)的理解,这里你的代码看起来像:

package com.example.game.bfs.alogrithm;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;


/**
   http://stackoverflow.com/questions/27768394/breadth-first-search-is-not-returning-the-shortest-path
 **/
public class BFS {

static class Cell {
    private int x;
    private int y;
    private String value;
    private boolean visitStatus;
    private Cell previousCell;

    public Cell(int x, int y, String value,boolean visitStatus) {
        this.x = x;
        this.y = y;
        this.value = value; 
        this.visitStatus=visitStatus;
    }

    public String toString()
    {
    return  "[" +   x + "," +   y + "]";
    }
}

private Cell[][] board;

private List<Cell> visited = new ArrayList<Cell>();

private boolean testDone;

public void setBoard(Cell[][] board) {
    this.board = board;
} 

public Cell getAdjacentUnvisitedCell(Cell cell)
{  
    int moves[][] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };
    // for diagonal moves :
    // int moves[][] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 }, {1,1} };
   for (int n = 0; n < moves.length ; ++n) {
       int ti = cell.x + moves[n][0];
       int tj = cell.y + moves[n][1];
      // System.out.println("ti,tj" + ti +"," + tj );  

       if (ti >= 0 && ti < board.length && tj >= 0 && tj < board[0].length) { 

          // System.out.println("getAdjacentUnvisitedCell : " + "[" + board[ti][tj].x + "," + board[ti][tj].y + "]" ); 
          // System.out.println("getAdjacentUnvisitedCell : board[ti][tj].visitStatus " + board[ti][tj].visitStatus ); 

           if(!board[ti][tj].visitStatus) {  
              return board[ti][tj];
           }
       }
   }  
   return null;  
} 

public void BFSearch(Cell start, Cell end) {  
   // BFS uses Queue data structure 
   Queue<Cell> q = new LinkedList<Cell>(); 
   Cell unVisitedadjCell = start; 
   Cell c =  null;
   q.add(start);
   visited.add(start);
   board[start.x][start.y].visitStatus = true;

   //printNode(start);

   while( !q.isEmpty() )
   { 
      c = q.peek(); 
      unVisitedadjCell = getAdjacentUnvisitedCell(c); 

      if(!testDone){
          testDone=true;  
      } 

      if ( unVisitedadjCell != null )
      {  visited.add(unVisitedadjCell); 
         board[unVisitedadjCell.x][unVisitedadjCell.y].visitStatus = true;

         printNode(unVisitedadjCell,c);
         unVisitedadjCell.previousCell=c; 
         q.add(unVisitedadjCell);
      }
      else
      {
         q.remove();
      }
   }
   System.out.println("Shortest path");

   while ( ( c != null ) && ( c!=start))
       {
       printNode(c.previousCell,c);
       c = c.previousCell;
       }

   visited.clear();     //Clear visited property of nodes
}


private void printNode(Cell c,Cell node) {
    System.out.println("For Node " + node.x +"," + node.y + ",  " + "Just Visited : " + c + " previous was " + node.previousCell);
} 

public static void main(String[] args) {
    Cell[][] cells = new Cell[10][10];
    for (int i = 0; i < 10; i++) {
        for (int j = 0; j < 10; j++) {
            cells[i][j] = new Cell(i, j, "defaultvalue",false);
        }
    }
    cells[0][1].value = "B";
    cells[0][2].value = "B";
    cells[1][1].value = "B";
    cells[1][3].value = "B";
    cells[2][1].value = "B";
    cells[2][2].value = "B";
    cells[2][3].value = "B";
    cells[2][7].value = "B";
    cells[2][8].value = "B"; 

    BFS board = new BFS();
    board.setBoard(cells);

    board.BFSearch(cells[0][0], cells[1][4]);
}


}

java -cp . java -cp。 com.example.game.bfs.alogrithm.BFS com.example.game.bfs.alogrithm.BFS

For Node 0,0,  Just Visited : [1,0] previous was null
For Node 0,0,  Just Visited : [0,1] previous was null
For Node 1,0,  Just Visited : [2,0] previous was [0,0]
For Node 1,0,  Just Visited : [1,1] previous was [0,0]
For Node 0,1,  Just Visited : [0,2] previous was [0,0]
For Node 2,0,  Just Visited : [3,0] previous was [1,0]
For Node 2,0,  Just Visited : [2,1] previous was [1,0]
For Node 1,1,  Just Visited : [1,2] previous was [1,0]
For Node 0,2,  Just Visited : [0,3] previous was [0,1]
For Node 3,0,  Just Visited : [4,0] previous was [2,0]
For Node 3,0,  Just Visited : [3,1] previous was [2,0]
For Node 2,1,  Just Visited : [2,2] previous was [2,0]
For Node 1,2,  Just Visited : [1,3] previous was [1,1]
For Node 0,3,  Just Visited : [0,4] previous was [0,2]
For Node 4,0,  Just Visited : [5,0] previous was [3,0]
For Node 4,0,  Just Visited : [4,1] previous was [3,0]
For Node 3,1,  Just Visited : [3,2] previous was [3,0]
For Node 2,2,  Just Visited : [2,3] previous was [2,1]
For Node 1,3,  Just Visited : [1,4] previous was [1,2]
For Node 0,4,  Just Visited : [0,5] previous was [0,3]
For Node 5,0,  Just Visited : [6,0] previous was [4,0]
For Node 5,0,  Just Visited : [5,1] previous was [4,0]
For Node 4,1,  Just Visited : [4,2] previous was [4,0]
For Node 3,2,  Just Visited : [3,3] previous was [3,1]
For Node 2,3,  Just Visited : [2,4] previous was [2,2]
For Node 1,4,  Just Visited : [1,5] previous was [1,3]
For Node 0,5,  Just Visited : [0,6] previous was [0,4]
For Node 6,0,  Just Visited : [7,0] previous was [5,0]
For Node 6,0,  Just Visited : [6,1] previous was [5,0]
For Node 5,1,  Just Visited : [5,2] previous was [5,0]
For Node 4,2,  Just Visited : [4,3] previous was [4,1]
For Node 3,3,  Just Visited : [3,4] previous was [3,2]
For Node 2,4,  Just Visited : [2,5] previous was [2,3]
For Node 1,5,  Just Visited : [1,6] previous was [1,4]
For Node 0,6,  Just Visited : [0,7] previous was [0,5]
For Node 7,0,  Just Visited : [8,0] previous was [6,0]
For Node 7,0,  Just Visited : [7,1] previous was [6,0]
For Node 6,1,  Just Visited : [6,2] previous was [6,0]
For Node 5,2,  Just Visited : [5,3] previous was [5,1]
For Node 4,3,  Just Visited : [4,4] previous was [4,2]
For Node 3,4,  Just Visited : [3,5] previous was [3,3]
For Node 2,5,  Just Visited : [2,6] previous was [2,4]
For Node 1,6,  Just Visited : [1,7] previous was [1,5]
For Node 0,7,  Just Visited : [0,8] previous was [0,6]
For Node 8,0,  Just Visited : [9,0] previous was [7,0]
For Node 8,0,  Just Visited : [8,1] previous was [7,0]
For Node 7,1,  Just Visited : [7,2] previous was [7,0]
For Node 6,2,  Just Visited : [6,3] previous was [6,1]
For Node 5,3,  Just Visited : [5,4] previous was [5,2]
For Node 4,4,  Just Visited : [4,5] previous was [4,3]
For Node 3,5,  Just Visited : [3,6] previous was [3,4]
For Node 2,6,  Just Visited : [2,7] previous was [2,5]
For Node 1,7,  Just Visited : [1,8] previous was [1,6]
For Node 0,8,  Just Visited : [0,9] previous was [0,7]
For Node 9,0,  Just Visited : [9,1] previous was [8,0]
For Node 8,1,  Just Visited : [8,2] previous was [8,0]
For Node 7,2,  Just Visited : [7,3] previous was [7,1]
For Node 6,3,  Just Visited : [6,4] previous was [6,2]
For Node 5,4,  Just Visited : [5,5] previous was [5,3]
For Node 4,5,  Just Visited : [4,6] previous was [4,4]
For Node 3,6,  Just Visited : [3,7] previous was [3,5]
For Node 2,7,  Just Visited : [2,8] previous was [2,6]
For Node 1,8,  Just Visited : [1,9] previous was [1,7]
For Node 9,1,  Just Visited : [9,2] previous was [9,0]
For Node 8,2,  Just Visited : [8,3] previous was [8,1]
For Node 7,3,  Just Visited : [7,4] previous was [7,2]
For Node 6,4,  Just Visited : [6,5] previous was [6,3]
For Node 5,5,  Just Visited : [5,6] previous was [5,4]
For Node 4,6,  Just Visited : [4,7] previous was [4,5]
For Node 3,7,  Just Visited : [3,8] previous was [3,6]
For Node 2,8,  Just Visited : [2,9] previous was [2,7]
For Node 9,2,  Just Visited : [9,3] previous was [9,1]
For Node 8,3,  Just Visited : [8,4] previous was [8,2]
For Node 7,4,  Just Visited : [7,5] previous was [7,3]
For Node 6,5,  Just Visited : [6,6] previous was [6,4]
For Node 5,6,  Just Visited : [5,7] previous was [5,5]
For Node 4,7,  Just Visited : [4,8] previous was [4,6]
For Node 3,8,  Just Visited : [3,9] previous was [3,7]
For Node 9,3,  Just Visited : [9,4] previous was [9,2]
For Node 8,4,  Just Visited : [8,5] previous was [8,3]
For Node 7,5,  Just Visited : [7,6] previous was [7,4]
For Node 6,6,  Just Visited : [6,7] previous was [6,5]
For Node 5,7,  Just Visited : [5,8] previous was [5,6]
For Node 4,8,  Just Visited : [4,9] previous was [4,7]
For Node 9,4,  Just Visited : [9,5] previous was [9,3]
For Node 8,5,  Just Visited : [8,6] previous was [8,4]
For Node 7,6,  Just Visited : [7,7] previous was [7,5]
For Node 6,7,  Just Visited : [6,8] previous was [6,6]
For Node 5,8,  Just Visited : [5,9] previous was [5,7]
For Node 9,5,  Just Visited : [9,6] previous was [9,4]
For Node 8,6,  Just Visited : [8,7] previous was [8,5]
For Node 7,7,  Just Visited : [7,8] previous was [7,6]
For Node 6,8,  Just Visited : [6,9] previous was [6,7]
For Node 9,6,  Just Visited : [9,7] previous was [9,5]
For Node 8,7,  Just Visited : [8,8] previous was [8,6]
For Node 7,8,  Just Visited : [7,9] previous was [7,7]
For Node 9,7,  Just Visited : [9,8] previous was [9,6]
For Node 8,8,  Just Visited : [8,9] previous was [8,7]
For Node 9,8,  Just Visited : [9,9] previous was [9,7]
Shortest path
For Node 9,9,  Just Visited : [9,8] previous was [9,8]
For Node 9,8,  Just Visited : [9,7] previous was [9,7]
For Node 9,7,  Just Visited : [9,6] previous was [9,6]
For Node 9,6,  Just Visited : [9,5] previous was [9,5]
For Node 9,5,  Just Visited : [9,4] previous was [9,4]
For Node 9,4,  Just Visited : [9,3] previous was [9,3]
For Node 9,3,  Just Visited : [9,2] previous was [9,2]
For Node 9,2,  Just Visited : [9,1] previous was [9,1]
For Node 9,1,  Just Visited : [9,0] previous was [9,0]
For Node 9,0,  Just Visited : [8,0] previous was [8,0]
For Node 8,0,  Just Visited : [7,0] previous was [7,0]
For Node 7,0,  Just Visited : [6,0] previous was [6,0]
For Node 6,0,  Just Visited : [5,0] previous was [5,0]
For Node 5,0,  Just Visited : [4,0] previous was [4,0]
For Node 4,0,  Just Visited : [3,0] previous was [3,0]
For Node 3,0,  Just Visited : [2,0] previous was [2,0]
For Node 2,0,  Just Visited : [1,0] previous was [1,0]
For Node 1,0,  Just Visited : [0,0] previous was [0,0]

Then try with commented code with diagonals... 然后尝试使用带有对角线的注释代码...

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