迷宫的java代码,java迷宫寻路
关于Java走迷宫的问题。我已经有相关代码了，但是我看不懂。麻烦高手帮忙注释一下，然后再修改点儿。
package 走迷宫;
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import java.awt.BorderLayout;
import java.awt.Color;
import java.awt.GridLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.TimerTask;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JOptionPane;
import javax.swing.JPanel;
// 迷宫
public class Maze extends JFrame implements ActionListener {
private JPanel panel;
private JPanel northPanel;
private JPanel centerPanel;
private MazeGrid grid[][];
private JButton restart;
private JButton dostart;
private int rows;// rows 和cols目前暂定只能是奇数
private int cols;
private ListString willVisit;
private ListString visited;
private LinkedListString comed;
private long startTime;
private long endTime;
public Maze() {
rows = 25;
cols = 25;
willVisit = new ArrayListString();
visited = new ArrayListString();
comed = new LinkedListString();
init();
this.setTitle("回溯法--走迷宫");
this.add(panel);
this.pack();
this.setVisible(true);
this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
public void init() {
panel = new JPanel();
northPanel = new JPanel();
centerPanel = new JPanel();
panel.setLayout(new BorderLayout());
restart = new JButton("重新生成迷宫");
dostart = new JButton("开始走迷宫");
grid = new MazeGrid[rows][cols];
centerPanel.setLayout(new GridLayout(rows, cols, 1, 1));
centerPanel.setBackground(new Color(0, 0, 0));
northPanel.add(restart);
northPanel.add(dostart);
dostart.addActionListener(this);
restart.addActionListener(this);
for (int i = 0; i grid.length; i++)
for (int j = 0; j grid[i].length; j++) {
if (j % 2 == 0 i % 2 == 0)
grid[i][j] = new MazeGrid(true, 20, 20);
else
grid[i][j] = new MazeGrid(false, 20, 20);
}
grid[0][0].setVisited(true);
grid[0][0].setPersonCome(true);
grid[0][0].setStart(true);
visited.add("0#0");
grid[rows - 1][cols - 1].setEnd(true);
grid = createMap(grid, 0, 0);
for (int i = 0; i grid.length; i++)
for (int j = 0; j grid[i].length; j++) {
grid[i][j].repaint();
centerPanel.add(grid[i][j]);
}
panel.add(northPanel, BorderLayout.NORTH);
panel.add(centerPanel, BorderLayout.CENTER);
}
/**
* 生成迷宫
*
* @param mazeGrid
* @param x
* @param y
* @return
*/
public MazeGrid[][] createMap(MazeGrid mazeGrid[][], int x, int y) {
int visitX = 0;
int visitY = 0;
if (x - 2 = 0) {
if (!mazeGrid[x - 2][y].isVisited()) {
willVisit.add((x - 2) + "#" + y);
}
}
if (x + 2 cols) {
if (!mazeGrid[x + 2][y].isVisited()) {
willVisit.add((x + 2) + "#" + y);
}
}
if (y - 2 = 0) {
if (!mazeGrid[x][y - 2].isVisited()) {
willVisit.add(x + "#" + (y - 2));
}
}
if (y + 2 rows) {
if (!mazeGrid[x][y + 2].isVisited()) {
willVisit.add(x + "#" + (y + 2));
}
}
if (!willVisit.isEmpty()) {
int visit = (int) (Math.random() * willVisit.size());
String id = willVisit.get(visit);
visitX = Integer.parseInt(id.split("#")[0]);
visitY = Integer.parseInt(id.split("#")[1]);
mazeGrid[(visitX + x) / 2][(visitY + y) / 2].setMark(true);
mazeGrid[visitX][visitY].setVisited(true);
if (!visited.contains(id)) {// 将这个点加到已访问中去
visited.add(id);
}
willVisit.clear();
createMap(mazeGrid, visitX, visitY);
} else {
if (!visited.isEmpty()) {
String id = visited.remove(visited.size() - 1);// 取出最后一个元素
visitX = Integer.parseInt(id.split("#")[0]);
visitY = Integer.parseInt(id.split("#")[1]);
mazeGrid[visitX][visitY].setVisited(true);
createMap(mazeGrid, visitX, visitY);
}
}
return mazeGrid;
}
/**
* 走迷宫
*
* @param mazeGrid
* @param x
* @param y
*/
public String goMaze(MazeGrid mazeGrid[][], int x, int y) {
int comeX = 0;
int comeY = 0;
// left
if (x - 1 = 0) {
if (mazeGrid[x - 1][y].isMark()) {
if (!comed.contains((x - 1) + "#" + y))
willVisit.add((x - 1) + "#" + y);
}
}
// right
if (x + 1 cols) {
if (mazeGrid[x + 1][y].isMark()) {
if (!comed.contains((x + 1) + "#" + y))
willVisit.add((x + 1) + "#" + y);
}
}
// up
if (y - 1 = 0) {
if (mazeGrid[x][y - 1].isMark()) {
if (!comed.contains(x + "#" + (y - 1)))
willVisit.add(x + "#" + (y - 1));
}
}
// down
if (y + 1 rows) {
if (mazeGrid[x][y + 1].isMark()) {
if (!comed.contains(x + "#" + (y + 1)))
willVisit.add(x + "#" + (y + 1));
}
}
if (!willVisit.isEmpty()) {
int visit = (int) (Math.random() * willVisit.size());
String id = willVisit.get(visit);
comeX = Integer.parseInt(id.split("#")[0]);
comeY = Integer.parseInt(id.split("#")[1]);
mazeGrid[x][y].setPersonCome(false);
mazeGrid[comeX][comeY].setPersonCome(true);
mazeGrid[x][y].repaint();
mazeGrid[comeX][comeY].repaint();
willVisit.clear();
comed.add(x + "#" + y);
} else {
if (!comed.isEmpty()) {
String id = comed.removeLast();
comeX = Integer.parseInt(id.split("#")[0]);
comeY = Integer.parseInt(id.split("#")[1]);
mazeGrid[x][y].setPersonCome(false);
mazeGrid[comeX][comeY].setPersonCome(true);
mazeGrid[x][y].repaint();
mazeGrid[comeX][comeY].repaint();
comed.addFirst(x + "#" + y);
}
}
return comeX + "#" + comeY;
}
int comeX = 0;
int comeY = 0;
public void actionPerformed(ActionEvent e) {
if (e.getActionCommand().equals("重新生成迷宫")) {
refreshMap(grid);
} else if (e.getActionCommand().equals("开始走迷宫")) {
startTime = System.currentTimeMillis();
dostart.setVisible(false);
restart.setText("禁止刷新");
int delay = 1000;
int period = 500;// 循环间隔
java.util.Timer timer = new java.util.Timer();
timer.scheduleAtFixedRate(new TimerTask() {
public void run() {
if (grid[rows - 1][cols - 1].isPersonCome()) {
endTime = System.currentTimeMillis();
JOptionPane.showMessageDialog(null, "已经走出迷宫，耗时"
+ (endTime - startTime) / 1000 + "秒", "消息提示",
JOptionPane.ERROR_MESSAGE);
this.cancel();
restart.setText("重新生成迷宫");
} else {
String id = goMaze(grid, comeX, comeY);
comeX = Integer.parseInt(id.split("#")[0]);
comeY = Integer.parseInt(id.split("#")[1]);
}
}
}, delay, period);
}
}
/**
* 刷新地图
*/
public void refreshMap(MazeGrid mazeGrid[][]) {
comeX = 0;
comeY = 0;
willVisit.clear();
visited.clear();
comed.clear();
this.remove(panel);
init();
this.add(panel);
this.pack();
this.setVisible(true);
}
public static void main(String args[]) {
long start = System.currentTimeMillis();
new Maze();
long end = System.currentTimeMillis();
System.out.println("使用ArrayList生成迷宫耗时：" + (end - start) + "毫秒");
}
}
请帮忙用数据结构（java版）的知识解决这道迷宫问题的程序代码。
我这是用c写的。你可以看看，希望能帮助到你。
#include"stdlib.h"
#include"stdio.h"
#define N 50
#define M 50
int X;
int maze[N+2][M+2];
struct point{
int row,col,predecessor;
}queue[512];
int head=0,tail=0;
void shoudong_maze(int m,int n){
int i,j;
printf("\n\n");
printf("请按行输入迷宫，0表示通路，1表示障碍:\n\n");
for(i=0;im;i++)
for(j=0;jn;j++)
scanf("%d",maze[i][j]);
}
void zidong_maze(int m,int n){
int i,j;
printf("\n迷宫生成中……\n\n");
system("pause");
for(i=0;im;i++)
for(j=0;jn;j++)
maze[i][j]=rand()%2;
//由于rand()产生的随机数是从0到RAND_MAX
//RAND_MAX是定义在stdlib.h中的,其值至少为32767)
//要产生从X到Y的数,只需要这样写：k=rand()%(Y-X+1)+X;
}
void print_maze(int m,int n){
int i,j;
printf("\n迷宫生成结果如下:\n\n");
printf("迷宫入口\n");
printf("↓");
for(i=0;im;i++)
{printf("\n");
for(j=0;jn;j++)
{if(maze[i][j]==0) printf("□");
if(maze[i][j]==1) printf("■");}
}
printf("→迷宫出口\n");
}
void result_maze(int m,int n)
{ int i,j;
printf("迷宫通路(用☆表示)如下所示：\n\t");
for(i=0;im;i++)
{ printf("\n");
for(j=0;jn;j++)
{if(maze[i][j]==0||maze[i][j]==2) printf("□");
if(maze[i][j]==1) printf("■");
if(maze[i][j]==3) printf("☆");
}
}
}
void enqueue(struct point p)
{ queue[tail]=p;
tail++;
}
struct point dequeue()
{ head++;
return queue[head-1];
}
int is_empty()
{ return head==tail;
}
void visit(int row,int col,int maze[52][52])
{ struct point visit_point={row,col,head-1};
maze[row][col]=2;
enqueue(visit_point);
}
int mgpath(int maze[52][52],int m,int n)
{ X=1;
struct point p={0,0,-1};
if(maze[p.row][p.col]==1)
{ printf("\n===============================================\n");
printf("此迷宫无解\n\n");X=0;return 0;}
maze[p.row][p.col]=2;
enqueue(p);
while(!is_empty())
{p=dequeue();
if((p.row==m-1)(p.col==n-1)) break;
if((p.col+1n)(maze[p.row][p.col+1]==0)) visit(p.row,p.col+1,maze);
if((p.row+1m)(maze[p.row+1][p.col]==0)) visit(p.row+1,p.col,maze);
if((p.col-1=0)(maze[p.row][p.col-1]==0)) visit(p.row,p.col-1,maze);
if((p.row-1=0)(maze[p.row-1][p.col]==0)) visit(p.row-1,p.col,maze);
}
if(p.row==m-1p.col==n-1)
{printf("\n==================================================================\n");
printf("迷宫路径为：\n");
printf("(%d,%d)\n",p.row,p.col);
maze[p.row][p.col]=3;
while(p.predecessor!=-1)
{p=queue[p.predecessor];
printf("(%d,%d)\n",p.row,p.col);
maze[p.row][p.col]=3;
}
}
else {printf("\n=============================================================\n");
printf("此迷宫无解！\n\n");X=0;}
return 0;
}
int main()
{int i,m,n,cycle=0;
while(cycle!=(-1))
{
printf("********************************************************************************\n");
printf(" ☆欢迎进入迷宫求解系统☆\n");
printf(" 设计者:尹旭 林静波（信息2班）\n");
printf("********************************************************************************\n");
printf(" 手动生成迷宫 请按：1\n");
printf(" 自动生成迷宫 请按：2\n");
printf(" 退出 请按：3\n\n");
printf("********************************************************************************\n");
printf("\n");
printf("请选择你的操作：\n");
scanf("%d",i);
switch(i)
{case 1:printf("\n请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
while((m=0||m50)||(n=0||n50))
{ printf("\n抱歉，你输入的行列数超出预设范围(0-50,0-50),请重新输入：\n\n");
printf("请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
}
shoudong_maze(m,n);
print_maze(m,n);
mgpath(maze,m,n);
if(X!=0)
result_maze(m,n);
printf("\n\nPress Enter Contiue!\n");
getchar();
while(getchar()!='\n');
break;
case 2:printf("\n请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
while((m=0||m50)||(n=0||n50))
{printf("\n抱歉，你输入的行列数超出预设范围(0-50,0-50),请重新输入：\n\n");
printf("请输入行数：");
scanf("%d",m);
printf("\n");
printf("请输入列数：");
scanf("%d",n);
}
zidong_maze(m,n);
print_maze(m,n);
mgpath(maze,m,n);
if(X!=0)
result_maze(m,n);
printf("\n\nPress Enter Contiue!\n");getchar();while(getchar()!='\n');break;
case 3:cycle=(-1);
break;
default:printf("\n");
printf("你的输入有误!\n");
printf("\nPress Enter Contiue!\n");
getchar();
while(getchar()!='\n');break;
}
}
}
JAVA迷宫问题，怎么输出所走路径的坐标
在Way函数的if(Maza[Loci][Locj]==0) 语句里面Maza[Loci][Locj]=2;前面加一句
System.out.println("The Maza route is ("+Loci+","+Locj+")");就打印出程序所走路径的坐标了.
求大佬讲解一下这段java代码
//直译了哦,只有片面的代码,只能靠推测,其他你自己补充下:
//这个主要是对一个图像控件设置了键盘侦听事件,按下键盘时候对其进行坐标移位操作
//必须知道:ImageView图像控件,显示图片或者其他;num是一个二位数组,这里可以理解成矩阵地图;
public void control(int[][] num, ImageView iv) {
//对控件设置键盘侦听,
iv.setOnKeyPressed(e - {
//设置一个变量,从下面使用看,这个应该是一个图像步进单位
int m = 30;
//以下为:try catch封装的键盘四向按键判断
try {
//字面意义推测:row应该是行,col应该是列,这是用二维数组画了一个矩阵地图,0不能走,1可以走;
//如果键盘按下的键是s,并且"行+1"小于地图中的行的个数,并且:地图[row+1行][col列]的值不等于0
if (e.getCode() == KeyCode.S row + 1 num.length num[row + 1][col] != 0) {
//就设置图像y坐标(在原有坐标的基础上进行一个步进单位的叠加)
iv.setY(iv.getY() + m);
//行递增1:row=row+1;
row++;
}
//其实这里与下面与上面都是一个道理,仅仅判断的条件与按键不同罢了
if (e.getCode() == KeyCode.W row - 1 = 0 num[row - 1][col] != 0) {
iv.setY(iv.getY() - m);
//行递减1
row--;
}
//同上
if (e.getCode() == KeyCode.A col - 1 = 0 num[row][col - 1] != 0) {
iv.setX(iv.getX() - m);
//列递减1
col--;
}
//同上
if (e.getCode() == KeyCode.D col + 1 num[0].length num[row][col + 1] != 0) {
iv.setX(iv.getX() + m);
//列递增1
col++;
}
}catch (Exception n) {}
//如果行等于地图"行"的最大的索引值,并且列等于第一行数组的最大索引值
if (row == num.length - 1 col == num[0].length - 1) {
//就调用win函数;
win();
}
});
}
急！！求用Java实现牛郎织女迷宫问题的代码
不是不想帮你。像这种问题 需要去专业性网站去问。并且 分数一定要高。
比如 这个论坛里 高手都往这里走。 再高高的高手只会在自己的领域。不会上网上论坛。嘿嘿
求走迷宫问题的算法，要求用Java写的？
public class Maze { private int[][] maze = null;
private int[] xx = { 1, 0, -1, 0 };
private int[] yy = { 0, 1, 0, -1 };
private Queue queue = null; public Maze(int[][] maze) {
this.maze = maze;
queue = new Queue(maze.length * maze.length);
} public void go() {
Point outPt = new Point(maze.length - 1, maze[0].length - 1);
Point curPt = new Point(0, 0);
Node curNode = new Node(curPt, null);
maze[curPt.x][curPt.y] = 2;
queue.entryQ(curNode); while (!queue.isEmpty()) {
curNode = queue.outQ();
for (int i = 0; i xx.length; ++i) {
Point nextPt = new Point();
nextPt.x = (curNode.point).x + xx[i];
nextPt.y = (curNode.point).y + yy[i];
if (check(nextPt)) {
Node nextNode = new Node(nextPt, curNode);
queue.entryQ(nextNode);
maze[nextPt.x][nextPt.y] = 2;
if (nextPt.equals(outPt)) {
java.util.StackNode stack = new java.util.StackNode();
stack.push(nextNode);
while ((curNode = nextNode.previous) != null) {
nextNode = curNode;
stack.push(curNode);
}
System.out.println("A Path is:");
while (!stack.isEmpty()) {
curNode = stack.pop();
System.out.println(curNode.point);
}
return;
}
}
}
}
System.out.println("Non solution!");
} private boolean check(Point p) {
if (p.x 0 || p.x = maze.length || p.y 0 || p.y = maze[0].length) {
return false;
}
if (maze[p.x][p.y] != 0) {
return false;
}
return true;
} public static void main(String[] args) {
int[][] maze = {
{ 0, 0, 1, 0, 1, 0, 1, 0, 1, 0 },
{ 0, 0, 1, 1, 1, 0, 0, 0, 1, 0 },
{ 0, 1, 0, 0, 1, 0, 0, 0, 0, 1 },
{ 0, 0, 0, 0, 1, 0, 0, 0, 1, 1 },
{ 0, 0, 1, 0, 0, 0, 0, 0, 0, 1 },
{ 1, 0, 1, 0, 1, 0, 0, 1, 0, 0 },
{ 0, 1, 0, 0, 1, 0, 0, 1, 0, 1 },
{ 1, 0, 1, 0, 1, 1, 0, 1, 0, 0 }
};
new Maze(maze).go();
} private class Queue { Node[] array = null;
int size = 0;
int len = 0;
int head = 0;
int tail = 0; public Queue(int n) {
array = new Node[n + 1];
size = n + 1;
} public boolean entryQ(Node node) {
if (isFull()) {
return false;
}
tail = (tail + 1) % size;
array[tail] = node;
len++;
return true;
} public Node outQ() {
if (isEmpty()) {
return null;
}
head = (head + 1) % size;
len--;
return array[head];
} public boolean isEmpty() {
return (len == 0 || head == tail) ? true : false;
} public boolean isFull() {
return ((tail + 1) % size == head) ? true : false;
}
} private class Node { Point point = null;
Node previous = null; public Node() {
this(null,null);
} public Node(Point point, Node node) {
this.point = point;
this.previous = node;
}
} private class Point { int x = 0;
int y = 0; public Point() {
this(0, 0);
} public Point(int x, int y) {
this.x = x;
this.y = y;
} public boolean equals(Point p) {
return (x == p.x) (y == p.y);
} @Override
public String toString() {
return "(" + x + "," + y + ")";
}
}
}
分享标题：迷宫的java代码,java迷宫寻路
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