1：往上流传的是双缓冲区的解法，但是这样会对内存照成浪费，并且如果在分布式应用中采用此方法，后果是很严重的

2：究其原因，是我们在repaint触发时，没有将当前绘制的图形进行显示

import java.awt.BorderLayout;

import java.awt.Color;

import java.awt.Graphics;

import java.awt.Graphics2D;

import java.awt.Point;

import java.awt.event.MouseAdapter;

import java.awt.event.MouseEvent;

import java.awt.event.MouseMotionAdapter;

import java.awt.geom.Line2D;

import java.util.Vector;

import javax.swing.JFrame;

import javax.swing.JPanel;

class LinePanel  extends JPanel{

private static final long serialVersionUID = 1L;

private Vector<Line2D> lines = new Vector<Line2D> ();//存储画板上所有的直线

private Line2D line = new Line2D.Double() ;

private Line2D cur = new Line2D.Double() ;//起点到当前光标的直线，临时的

public LinePanel(){

super();

this.setBackground(new Color(255,255,255));