求java版汉诺塔的演示程序

发布网友发布时间：2022-04-30 16:11

共3个回答

热心网友时间：2022-06-27 03:59

源代码：
/**
*本程序完成的功能是利用汉递规算法实现汉诺塔的动态演示程序
*/
import javax.swing.*;
import java.awt.geom.*;
import java.awt.event.*;
import java.awt.*;
public class Hanio extends JApplet implements ActionListener, Runnable
{
/**
*diskNum是盘子的数量
*/
private int diskNum ;
/**
*各个组件的句柄
*/
private JButton begin, stop;
private JLabel lDiskNum;
private JTextField text;
JPanel pane;
/**
*定义一个线程句柄
*/
private Thread animate;
/**
*定义a,b,c三个柱子上是否有盘子，有哪些盘子
*/
private int adisk[];
private int bdisk[];
private int cdisk[];
public void init()
{

  Container content = getContentPane();
  content.setLayout(new BorderLayout());
  lDiskNum = new JLabel(盘子的数目);

  text = new JTextField(8);

  begin = new JButton(开始);
  begin.addActionListener(this);

  stop = new JButton(停止);
  stop.addActionListener(this);

  pane = new JPanel();
  pane.setLayout(new FlowLayout());
  pane.add(lDiskNum);
  pane.add(text);
  pane.add(begin);
  pane.add(stop);
  content.add(pane, BorderLayout.SOUTH);

}
public void paint(Graphics g)
{
  Graphics2D g2D = (Graphics2D)g;
  Ellipse2D.Double ellipse;
  g2D.setPaint(getBackground());
  if(adisk != null)
  {
   /**
*消除以前画的盘子
*/
   for(int j=adisk.length, i=0; --j>=0; i++ )
   {
    ellipse = new Ellipse2D.Double(20+i*5, 180-i*10, 180-i*10, 20);
    g2D.fill(ellipse);
    ellipse = new Ellipse2D.Double(220+i*5, 180-i*10, 180-i*10, 20);
    g2D.fill(ellipse);
    ellipse = new Ellipse2D.Double(420+i*5, 180-i*10, 180-i*10, 20);
    g2D.fill(ellipse);

   }
   drawEllipse(g, 20, adisk);//画A组盘子
   drawEllipse(g, 220, bdisk);//画B组盘子
   drawEllipse(g, 420, cdisk);//画C组盘子

  }
  pane.repaint();
}
public void update(Graphics g)
{
  paint(g);
}
/**画出椭圆代表盘子，g是图形环境，x是最下面的盘子的横坐标，
*arr是柱子数组
*/
public void drawEllipse(Graphics g,int x,int arr[])
{
  Graphics2D g2D = (Graphics2D)g;
  Ellipse2D.Double ellipse;
  g2D.setPaint(Color.gray);
  g2D.draw(new Line2D.Double(x+90, 10, x+90, 180));
  for(int j=arr.length, i=0; --j>=0; i++ )
   if(arr[j] != 0)
   {
    if(i%2 == 0)
     g2D.setPaint(Color.blue);
    else
     g2D.setPaint(Color.red);
    ellipse = new Ellipse2D.Double(x+i*5, 180-i*10, 180-i*10, 20);
    g2D.fill(ellipse);
   }
}
public void actionPerformed(ActionEvent e)
{
  String command = e.getActionCommand();
  if(command.equals(开始))
  {
   /**
*进行初始化，开始的时候只有a柱子上有盘子，其他柱子都没有
*/
   diskNum = Integer.parseInt(text.getText());

   adisk = new int[diskNum];
   for(int i=0; i<adisk.length; i++)
    adisk[i] = 1;
   bdisk = new int[diskNum];
   for(int k=0; k<bdisk.length; k++)
    bdisk[k] = 0;
   cdisk = new int[diskNum];
   for(int i=0; i<cdisk.length; i++)
    cdisk[i] = 0;
   repaint();
   if(animate == null || !animate.isAlive())//创建一个线程
   {
    animate = new Thread(this);
    animate.start();
   }
  }
  if(command.equals(停止))
  {
   for(int k=0; k<bdisk.length; k++)
    bdisk[k] = 0;
   for(int i=0; i<cdisk.length; i++)
    cdisk[i] = 0;
   repaint();
   text.setText();
   animate = null;
  }
}
/**
*线程方法，在此调用汉诺塔执行移动盘子操作
*/
public void run()
{
  hanio(diskNum, 'A', 'B', 'C');
  repaint();
}
/**
*汉诺塔递规调用程序，n是盘子的数量，A，B，C分别代表三个柱子
*/
public void hanio(int n, char A, char B, char C)
{
  if(n > 1)
  {
   hanio(n-1, A, C, B);
   pause();//停顿几秒在执行
   switch(A)
   {
    case 'A':adisk[n-1] = 0;break;
    case 'B':bdisk[n-1] = 0;break;
    case 'C':cdisk[n-1] = 0;break;
    default:break;
   }
   switch(C)
   {
    case 'A':adisk[n-1] = 1;break;
    case 'B':bdisk[n-1] = 1;break;
    case 'C':cdisk[n-1] = 1;break;
    default:break;
   }
   repaint();
   hanio(n-1, B, A, C);
  }
  pause();
switch(A)
  {
   case 'A':adisk[n-1] = 0;break;
   case 'B':bdisk[n-1] = 0;break;
   case 'C':cdisk[n-1] = 0;break;
   default:break;
  }
  switch(C)
  {
   case 'A':adisk[n-1] = 1;break;
   case 'B':bdisk[n-1] = 1;break;
   case 'C':cdisk[n-1] = 1;break;
   default:break;
  }
  repaint();

}

/**
*每隔半妙钟移动一个盘子
*/
public void pause()
{
  try{
   Thread.sleep(500);//可以修改此值加快盘子移动的速度
  }catch(InterruptedException e){}
}
}

热心网友时间：2022-06-27 04:00

根本用不着这么麻烦
用递归就可以了，虽然内存占用大了一点。
该问题的基本思路是将A柱的盘子借助B移动到C上去。
当A柱上的盘子为1时只需1步，
由此建立递归关系。

或者你心情好，用遍历一遍。

热心网友时间：2022-06-27 04:00

var vEngine;
var oEngine = frm.engine;
for (var i = 0; i < oEngine.length; i++)
{
if (oEngine[i].checked)
{
vEngine = oEngine[i].value;
break;
}
}
var url;
switch(vEngine)
{
case "google":
url = "http://images.google.com/images?q=" + kw + "&inlang=zh-CN&ie=GB2312&oe=GB2312";
break;
case "sina":
url = "http://pic.sina.com.cn/cgi-bin/retr/search?query=" + kw;
break;
case "":
url = "http://image.baidu.com/i?z=&s=1&ct=201326592&cl=2&lm=-1&tn=image&word=" + kw;
break;
case "ys":
url = "http://image.yisou.com/search?p=" + kw;
break;
default：

}
document.getElementById("search").src = url;
return false;
}
//-->
</Script>
</HEAD>

<BODY style="text-align: center; margin: 4" scroll="no" onload="document.all.keyword.focus();">
<form action="" method="GET" onsubmit="return frmCheck(this);">
<fieldset><legend>Power Search <sup>©</sup></legend>
<table>
<tr>
<td><input name="engine" type="radio" value="google" checked> GOOGLE</td>
<td><input name="engine" type="radio" value="sina"> SINA</td>
<td><input name="engine" type="radio" value=""> 百度</td>
<td><input name="engine" type="radio" value="ys"> 一搜</td>
</tr>
<tr>
<td colspan="4"><input type="text" name="keyword" size="30"> <input type="submit" value="搜索(S)"></td>
</tr>
</table>
</fieldset>
</form>
<iframe name="search" id="search" width="100%" height="100%" scrolling="inherit" frameborder="0"></iframe>