Maze problem (BFS)

1 Importjava.util.LinkedList;2 ImportJava.util.Queue;3 ImportJava.util.Stack;4 5  Public classBFS6 {7     Private Static Final intm=3;//Representative Row8     Private Static Final intn=3;//represents a column9     Ten     intDestx=1; One     intDesty=0; A      -     //int[][] maze={{0,0,1,1},{0,0,0,1},{1,1,0,1},{0,0,0,0}};//Maze Layout, 1 means obstacle -     int[] maze={{0,0,0},{0,1,0},{0,0,0}};//Maze Layout, 1 means obstacle the     int[] visit=New intM [N];//whether the tag has been visited -     int[] steparr={{-1,0},{1,0},{0,-1},{0,1}};//indicates the direction of walking: up or down -      -      +      -     classNode +     { A         intx, y;//represents the coordinates of the current position at         intStep; -         intPrex,prey; -          -Node (intXintYintPreX,intPreY,intStep) -         { -              This. x=x;//represents the horizontal axis of the current position in              This. y=y;//represents the ordinate of the current position -              This. Prex=prex;//represents the horizontal axis of the previous step in the current position to              This. Prey=prey;//An ordinate that represents the previous step in the current position +              This. Step=step;//It means walking away. -         }         the     } *      $      Public BooleanBFS ()Panax Notoginseng     { -Node node=NewNode (0,0,-1,-1,0);//Start Position theQueue<node> queue=NewLinkedlist<node> ();//the LinkedList class implements the queue interface +Stack<node> stack=NewStack<node>(); A          theQueue.offer (node);//Insert Node +          -          while(!queue.isempty ()) $         { $Node Head=queue.poll ();//Launch Queue Header -              -Stack.push (head);//used to save paths theVISIT[HEAD.X][HEAD.Y] = 1;//indicates that the current location has been accessed to prevent two coordinates from being accessed in a loop -              for(inti = 0; I < 4; i++) {//4 here means 4 directions.Wuyi                 intx = head.x + steparr[i][0];  the                 inty = head.y + steparr[i][1];  -                 //Exit Wu                 if(x ==destx && y = = Desty && maze[x][y] = = 0 && visit[x][y] = = 0) {//means to find a destination -                     //Print Path AboutNode top =Stack.pop ();  $System.out.println ("Steps:" + (Top.step + 1));//The path shown here is the shortest path, because it is the queue that holds every possible -SYSTEM.OUT.PRINTLN ("The Path:");  -System.out.println ((M-1) + "," + (N-1));  -System.out.println (top.x + "," +top.y);  A                     intPreX =Top.prex;  +                     intPreY =Top.prey;  the                      while(!Stack.isempty ()) {   -top =Stack.pop ();  $                         if(PreX = = top.x && PreY = = top.y) {//because every possible time is saved in the queue, it needs to be validated with the previously saved point theSystem.out.println (PreX + "," +PreY);  thePreX =Top.prex;  thePreY =Top.prey;  the                         }   -                            in                     }   the                     return true;  the                 }   About                 //BFS the                 if(x >= 0 && x < M && y >= 0 && y < N &&maze[x][y] = 0 && Visit[x][y] = = 0){   theNode NewNode =NewNode (x, Y, head.x, Head.y, Head.step + 1);  the Queue.offer (NewNode);  +                 }   -             } the                         Bayi         }         the         return false;  the     } -      -      the      Public Static voidMain (string[] args) the     { theBFS b=NewBFS (); the         if(B.BFS () = =false) -         { theSystem.out.println ("fail!"); the         } the     }94      the      the      the     98}

The shortest path of labyrinth can be obtained by using BFS.

If you use DFS, you can only judge if you can get out of the maze, and not know whether the path is the shortest path.

Maze problem (BFS)