上机编程题-迷宫问题

来源：互联网发布：端口能ping吗编辑：程序博客网时间：2024/06/06 08:42

用Java实现迷宫求解，基本实现就是DFS搜索，可以实现最短路径。 Point 是一个java类,主要属性石x,y，实现对坐标的抽象。

public final class Demo{private static int dirx[]={0,1,0,-1};private static int diry[]={1,0,-1,0};private int[][] flag;private Point[][] record;        //保存路径的栈    private Stack<Point> stack = new Stack<Point>();        public Demo()    {            }      /*    功能:从一个迷宫走出的最短路徑            输入:        一个N*M的数组,int[][] maze迷宫图作为输入，如        {0, 1, 0, 0, 0},        {0, 1, 0, 1, 0},         {0, 0, 0, 0, 0},         {0, 1, 1, 1, 0},         {0, 0, 0, 1, 0}};            输出:从左上角到右下角的最短路线：(0, 0)(1, 0)(2, 0)(2, 1)(2, 2)(2, 3)(2, 4)(3, 4)(4, 4)           */    public Stack<Point> go(int[][] maze)    {        Point out = new Point(maze.length - 1, maze[0].length - 1); //出口        Point in = new Point(0, 0); //入口//searchQueue<Point> queue= new LinkedList<Point>();queue.offer(in);Point current_point = new Point();Point new_point;flag = new int[maze.length][maze[0].length];  //record 走过没有record = new Point[maze.length][maze[0].length]; //record 前驱节点for(int i = 0 ; i < flag.length ; i++){for(int j = 0 ; j <  flag[0].length;j++){flag[i][j] = 0;}}flag[in.getX()][in.getY()] = 1;int x,y;while(!queue.isEmpty()){current_point = queue.peek();queue.poll();if(current_point.getX() == out.getX() && current_point.getY() == out.getY()){break;}for(int i=0;i<4;i++){ //move to 4 directionx = current_point.getX() + dirx[i];    y = current_point.getY() + diry[i];if(x >= in.getX() && x <= out.getX() && y >= in.getY() && y <= out.getY() && maze[x][y] != 1 && flag[x][y] == 0)// judge whether is accessible  or back or out of board  //防止坐标溢出，同时判断是否碰到墙壁或者走回原路{new_point = new Point(x,y);queue.add(new_point);flag[new_point.getX()][new_point.getY()] = 1;//recordrecord[new_point.getX()][new_point.getY()] = current_point;}}}//pathif(flag[current_point.getX()][current_point.getY()] != 0){Stack<Point> path = new Stack<Point>();path.push(out);int i = record[out.getX()][out.getY()].getX();int j = record[out.getX()][out.getY()].getY();while(i != in.getX() || j != in.getY()){path.push(new Point(i,j));int m = record[i][j].getX();int n = record[i][j].getY();i=m;j=n;}path.push(new Point(0,0));Collections.reverse(path);return path;}else{return null;}    }    }

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