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java霍夫算法源代码,java霍夫曼编码
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高分求java的RSA 和IDEA 加密解密算法
RSA算法非常简单，概述如下：
找两素数p和q
取n=p*q
取t=(p-1)*(q-1)
取任何一个数e,要求满足et并且e与t互素（就是最大公因数为1）
取d*e%t==1
这样最终得到三个数： n d e
设消息为数M (M n)
设c=(M**d)%n就得到了加密后的消息c
设m=(c**e)%n则 m == M，从而完成对c的解密。
注：**表示次方,上面两式中的d和e可以互换。
在对称加密中：
n d两个数构成公钥，可以告诉别人；
n e两个数构成私钥，e自己保留，不让任何人知道。
给别人发送的信息使用e加密，只要别人能用d解开就证明信息是由你发送的，构成了签名机制。
别人给你发送信息时使用d加密，这样只有拥有e的你能够对其解密。
rsa的安全性在于对于一个大数n，没有有效的方法能够将其分解
从而在已知n d的情况下无法获得e；同样在已知n e的情况下无法
求得d。
二实践
接下来我们来一个实践，看看实际的操作：
找两个素数：
p=47
q=59
这样
n=p*q=2773
t=(p-1)*(q-1)=2668
取e=63，满足et并且e和t互素
用perl简单穷举可以获得满主 e*d%t ==1的数d：
C:\Tempperl -e "foreach $i (1..9999){ print($i),last if $i*63%2668==1 }"
847
即d＝847
最终我们获得关键的
n=2773
d=847
e=63
取消息M=244我们看看
加密：
c=M**d%n = 244**847%2773
用perl的大数计算来算一下：
C:\Tempperl -Mbigint -e "print 244**847%2773"
465
即用d对M加密后获得加密信息c＝465
解密：
我们可以用e来对加密后的c进行解密，还原M：
m=c**e%n=465**63%2773 ：
C:\Tempperl -Mbigint -e "print 465**63%2773"
244
即用e对c解密后获得m=244 , 该值和原始信息M相等。
三字符串加密
把上面的过程集成一下我们就能实现一个对字符串加密解密的示例了。
每次取字符串中的一个字符的ascii值作为M进行计算，其输出为加密后16进制
的数的字符串形式，按3字节表示，如01F
代码如下：
#!/usr/bin/perl -w
#RSA 计算过程学习程序编写的测试程序
#watercloud 2003-8-12
#
use strict;
use Math::BigInt;
my %RSA_CORE = (n=2773,e=63,d=847); #p=47,q=59
my $N=new Math::BigInt($RSA_CORE{n});
my $E=new Math::BigInt($RSA_CORE{e});
my $D=new Math::BigInt($RSA_CORE{d});
print "N=$N D=$D E=$E\n";
sub RSA_ENCRYPT
{
my $r_mess = shift @_;
my ($c,$i,$M,$C,$cmess);
for($i=0;$i length($$r_mess);$i++)
{
$c=ord(substr($$r_mess,$i,1));
$M=Math::BigInt-new($c);
$C=$M-copy(); $C-bmodpow($D,$N);
$c=sprintf "%03X",$C;
$cmess.=$c;
}
return \$cmess;
}
sub RSA_DECRYPT
{
my $r_mess = shift @_;
my ($c,$i,$M,$C,$dmess);
for($i=0;$i length($$r_mess);$i+=3)
{
$c=substr($$r_mess,$i,3);
$c=hex($c);
$M=Math::BigInt-new($c);
$C=$M-copy(); $C-bmodpow($E,$N);
$c=chr($C);
$dmess.=$c;
}
return \$dmess;
}
my $mess="RSA 娃哈哈哈～～～";
$mess=$ARGV[0] if @ARGV = 1;
print "原始串：",$mess,"\n";
my $r_cmess = RSA_ENCRYPT(\$mess);
print "加密串：",$$r_cmess,"\n";
my $r_dmess = RSA_DECRYPT($r_cmess);
print "解密串：",$$r_dmess,"\n";
#EOF
测试一下：
C:\Tempperl rsa-test.pl
N=2773 D=847 E=63
原始串：RSA 娃哈哈哈～～～
加密串：5CB6CD6BC58A7709470AA74A0AA74A0AA74A6C70A46C70A46C70A4
解密串：RSA 娃哈哈哈～～～
C:\Tempperl rsa-test.pl 安全焦点（xfocus）
N=2773 D=847 E=63
原始串：安全焦点（xfocus）
加密串：3393EC12F0A466E0AA9510D025D7BA0712DC3379F47D51C325D67B
解密串：安全焦点（xfocus）
四提高
前面已经提到，rsa的安全来源于n足够大，我们测试中使用的n是非常小的，根本不能保障安全性，
我们可以通过RSAKit、RSATool之类的工具获得足够大的N 及D E。
通过工具，我们获得1024位的N及D E来测试一下：
n=0x328C74784DF31119C526D18098EBEBB943B0032B599CEE13CC2BCE7B5FCD15F90B66EC3A85F5005D
BDCDED9BDFCB3C4C265AF164AD55884D8278F791C7A6BFDAD55EDBC4F017F9CCF1538D4C2013433B383B
47D80EC74B51276CA05B5D6346B9EE5AD2D7BE7ABFB36E37108DD60438941D2ED173CCA50E114705D7E2
BC511951
d=0x10001
e=0xE760A3804ACDE1E8E3D7DC0197F9CEF6282EF552E8CEBBB7434B01CB19A9D87A3106DD28C523C2995
4C5D86B36E943080E4919CA8CE08718C3B0930867A98F635EB9EA9200B25906D91B80A47B77324E66AFF2
C4D70D8B1C69C50A9D8B4B7A3C9EE05FFF3A16AFC023731D80634763DA1DCABE9861A4789BD782A592D2B
1965
设原始信息
M=0x11111111111122222222222233333333333
完成这么大数字的计算依赖于大数运算库，用perl来运算非常简单：
A) 用d对M进行加密如下：
c=M**d%n :
C:\Tempperl -Mbigint -e " $x=Math::BigInt-bmodpow(0x11111111111122222222222233
333333333, 0x10001, 0x328C74784DF31119C526D18098EBEBB943B0032B599CEE13CC2BCE7B5F
CD15F90B66EC3A85F5005DBDCDED9BDFCB3C4C265AF164AD55884D8278F791C7A6BFDAD55EDBC4F0
17F9CCF1538D4C2013433B383B47D80EC74B51276CA05B5D6346B9EE5AD2D7BE7ABFB36E37108DD6
0438941D2ED173CCA50E114705D7E2BC511951);print $x-as_hex"
0x17b287be418c69ecd7c39227ab681ac422fcc84bb35d8a632543b304de288a8d4434b73d2576bd
45692b007f3a2f7c5f5aa1d99ef3866af26a8e876712ed1d4cc4b293e26bc0a1dc67e247715caa6b
3028f9461a3b1533ec0cb476441465f10d8ad47452a12db0601c5e8beda686dd96d2acd59ea89b91
f1834580c3f6d90898
即用d对M加密后信息为：
c=0x17b287be418c69ecd7c39227ab681ac422fcc84bb35d8a632543b304de288a8d4434b73d2576bd
45692b007f3a2f7c5f5aa1d99ef3866af26a8e876712ed1d4cc4b293e26bc0a1dc67e247715caa6b
3028f9461a3b1533ec0cb476441465f10d8ad47452a12db0601c5e8beda686dd96d2acd59ea89b91
f1834580c3f6d90898
B) 用e对c进行解密如下：
m=c**e%n ：
C:\Tempperl -Mbigint -e " $x=Math::BigInt-bmodpow(0x17b287be418c69ecd7c39227ab
681ac422fcc84bb35d8a632543b304de288a8d4434b73d2576bd45692b007f3a2f7c5f5aa1d99ef3
866af26a8e876712ed1d4cc4b293e26bc0a1dc67e247715caa6b3028f9461a3b1533ec0cb4764414
65f10d8ad47452a12db0601c5e8beda686dd96d2acd59ea89b91f1834580c3f6d90898, 0xE760A
3804ACDE1E8E3D7DC0197F9CEF6282EF552E8CEBBB7434B01CB19A9D87A3106DD28C523C29954C5D
86B36E943080E4919CA8CE08718C3B0930867A98F635EB9EA9200B25906D91B80A47B77324E66AFF
2C4D70D8B1C69C50A9D8B4B7A3C9EE05FFF3A16AFC023731D80634763DA1DCABE9861A4789BD782A
592D2B1965, 0x328C74784DF31119C526D18098EBEBB943B0032B599CEE13CC2BCE7B5FCD15F90
B66EC3A85F5005DBDCDED9BDFCB3C4C265AF164AD55884D8278F791C7A6BFDAD55EDBC4F017F9CCF
1538D4C2013433B383B47D80EC74B51276CA05B5D6346B9EE5AD2D7BE7ABFB36E37108DD60438941
D2ED173CCA50E114705D7E2BC511951);print $x-as_hex"
0x11111111111122222222222233333333333
(我的P4 1.6G的机器上计算了约5秒钟）
得到用e解密后的m=0x11111111111122222222222233333333333 == M
C) RSA通常的实现
RSA简洁幽雅，但计算速度比较慢，通常加密中并不是直接使用RSA 来对所有的信息进行加密，
最常见的情况是随机产生一个对称加密的密钥，然后使用对称加密算法对信息加密，之后用
RSA对刚才的加密密钥进行加密。
最后需要说明的是，当前小于1024位的N已经被证明是不安全的
自己使用中不要使用小于1024位的RSA，最好使用2048位的。
----------------------------------------------------------
一个简单的RSA算法实现JAVA源代码：
filename:RSA.java
/*
* Created on Mar 3, 2005
*
* TODO To change the template for this generated file go to
* Window - Preferences - Java - Code Style - Code Templates
*/
import java.math.BigInteger;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.FileWriter;
import java.io.FileReader;
import java.io.BufferedReader;
import java.util.StringTokenizer;
/**
* @author Steve
*
* TODO To change the template for this generated type comment go to
* Window - Preferences - Java - Code Style - Code Templates
*/
public class RSA {
/**
* BigInteger.ZERO
*/
private static final BigInteger ZERO = BigInteger.ZERO;
/**
* BigInteger.ONE
*/
private static final BigInteger ONE = BigInteger.ONE;
/**
* Pseudo BigInteger.TWO
*/
private static final BigInteger TWO = new BigInteger("2");
private BigInteger myKey;
private BigInteger myMod;
private int blockSize;
public RSA (BigInteger key, BigInteger n, int b) {
myKey = key;
myMod = n;
blockSize = b;
}
public void encodeFile (String filename) {
byte[] bytes = new byte[blockSize / 8 + 1];
byte[] temp;
int tempLen;
InputStream is = null;
FileWriter writer = null;
try {
is = new FileInputStream(filename);
writer = new FileWriter(filename + ".enc");
}
catch (FileNotFoundException e1){
System.out.println("File not found: " + filename);
}
catch (IOException e1){
System.out.println("File not found: " + filename + ".enc");
}
/**
* Write encoded message to 'filename'.enc
*/
try {
while ((tempLen = is.read(bytes, 1, blockSize / 8)) 0) {
for (int i = tempLen + 1; i bytes.length; ++i) {
bytes[i] = 0;
}
writer.write(encodeDecode(new BigInteger(bytes)) + " ");
}
}
catch (IOException e1) {
System.out.println("error writing to file");
}
/**
* Close input stream and file writer
*/
try {
is.close();
writer.close();
}
catch (IOException e1) {
System.out.println("Error closing file.");
}
}
public void decodeFile (String filename) {
FileReader reader = null;
OutputStream os = null;
try {
reader = new FileReader(filename);
os = new FileOutputStream(filename.replaceAll(".enc", ".dec"));
}
catch (FileNotFoundException e1) {
if (reader == null)
System.out.println("File not found: " + filename);
else
System.out.println("File not found: " + filename.replaceAll(".enc", "dec"));
}
BufferedReader br = new BufferedReader(reader);
int offset;
byte[] temp, toFile;
StringTokenizer st = null;
try {
while (br.ready()) {
st = new StringTokenizer(br.readLine());
while (st.hasMoreTokens()){
toFile = encodeDecode(new BigInteger(st.nextToken())).toByteArray();
System.out.println(toFile.length + " x " + (blockSize / 8));
if (toFile[0] == 0 toFile.length != (blockSize / 8)) {
temp = new byte[blockSize / 8];
offset = temp.length - toFile.length;
for (int i = toFile.length - 1; (i = 0) ((i + offset) = 0); --i) {
temp[i + offset] = toFile[i];
}
toFile = temp;
}
/*if (toFile.length != ((blockSize / 8) + 1)){
temp = new byte[(blockSize / 8) + 1];
System.out.println(toFile.length + " x " + temp.length);
for (int i = 1; i temp.length; i++) {
temp[i] = toFile[i - 1];
}
toFile = temp;
}
else
System.out.println(toFile.length + " " + ((blockSize / 8) + 1));*/
os.write(toFile);
}
}
}
catch (IOException e1) {
System.out.println("Something went wrong");
}
/**
* close data streams
*/
try {
os.close();
reader.close();
}
catch (IOException e1) {
System.out.println("Error closing file.");
}
}
/**
* Performs ttbase/tt^supttpow/tt/sup within the modular
* domain of ttmod/tt.
*
* @param base the base to be raised
* @param pow the power to which the base will be raisded
* @param mod the modular domain over which to perform this operation
* @return ttbase/tt^supttpow/tt/sup within the modular
* domain of ttmod/tt.
*/
public BigInteger encodeDecode(BigInteger base) {
BigInteger a = ONE;
BigInteger s = base;
BigInteger n = myKey;
while (!n.equals(ZERO)) {
if(!n.mod(TWO).equals(ZERO))
a = a.multiply(s).mod(myMod);
s = s.pow(2).mod(myMod);
n = n.divide(TWO);
}
return a;
}
}
在这里提供两个版本的RSA算法JAVA实现的代码下载：
1. 来自于 的RSA算法实现源代码包:
2. 来自于 的实现:
- 源代码包
- 编译好的jar包
另外关于RSA算法的php实现请参见文章:
php下的RSA算法实现
关于使用VB实现RSA算法的源代码下载(此程序采用了psc1算法来实现快速的RSA加密):
RSA加密的JavaScript实现:
相关问答
Q1: JAVA 求助一段算法代码！求大神~~~~~
看完你的，自己写了一个。很简陋。你的改动比较大。一时半会改不了。
你的写好了。改动有点大。鼠标事件mousePressed()中实现移动。由于时间没做优化，主要处理方法是判断当前listenerPanel的上下左右是否存在上面是0的listenerPanel，存在则交换上面数字及背景颜色。自己可以优化下里面代码，
思路：
PuzzleTile jb = (PuzzleTile) e.getSource();
for(int i=0;ilistenerPanel.length;i++){
if(jb.equals(listenerPanel[i])){
//判断当前listenerPanel[i]上下左右是否存有listenerPanel的上面数字是0的，如果存在
则把当前的listenerPanel[i]的背景颜色及数字与上面是0 的交换。判断周围是否存在有点及是否交换有点复杂。
}
}
代码修改如下：少量注释
import java.awt.Color;
import java.awt.Font;
import java.awt.FontMetrics;
import java.awt.Graphics;
import javax.swing.*;
public class PuzzleTile extends JPanel{
private String tileNumber;
public PuzzleTile(int number) {
super();
if (number == 0) {
this.setBackground(Color.white);
}
else {
this.setBackground(Color.darkGray);
}
this.tileNumber = "" + number;
}
public void setTitleNumber(int tileNumber){//设置上面的数字
this.tileNumber=tileNumber+"";
}
public int getTitleNumber(){//获得上面的数字
return Integer.parseInt(tileNumber);
}
public void paintComponent(Graphics graphics) {
Font a=new Font("Arial",Font.BOLD,30);
graphics.setFont(a);
graphics.setColor(Color.white);
super.paintComponent(graphics);
FontMetrics b=graphics.getFontMetrics(a);
int c=b.stringWidth(tileNumber);
int d=b.getAscent();
int e=getWidth()/2-c/2;
int f=getHeight()/2+d/2;
graphics.drawString(tileNumber,e,f);
}
}
import java.awt.*;
import javax.swing.*;
import java.awt.event.*;
import java.util.ArrayList;
import java.util.Random;
public class SlidingPuzzle extends JFrame implements MouseListener
{
public static void main(String[] args){
SlidingPuzzle frame=new SlidingPuzzle();
frame.TestPanel();
frame.setTitle("Numeric Sliding Puzzle");
frame.setSize(400,400);
frame.setVisible(true);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
PuzzleTile[] listenerPanel;
public void TestPanel(){
Container container=getContentPane();
container.setLayout(new GridLayout(3,3,5,5));
listenerPanel=new PuzzleTile[9];
ArrayListInteger myList=new ArrayListInteger();
int m;
for(int i=0;i9;i++){
m=new Random().nextInt(9);
if(!myList.contains(m))
myList.add(m);
else
i--;
}
for(int i=0;ilistenerPanel.length;i++){
listenerPanel[i]=new PuzzleTile(myList.get(i));
container.add(listenerPanel[i]);
listenerPanel[i].addMouseListener(this);
}
}
public void mousePressed(MouseEvent e){
PuzzleTile jb = (PuzzleTile) e.getSource();
int m=jb.getTitleNumber();
//依次判断每一个listenerPanel上下左右是否存在上面数字为0的listenerPanel
if(jb.equals(listenerPanel[0])){
if(listenerPanel[1].getTitleNumber()==0){
listenerPanel[0].setBackground(Color.white);
listenerPanel[0].setTitleNumber(0);
listenerPanel[1].setTitleNumber(m);
listenerPanel[1].setBackground(Color.darkGray);
}
if(listenerPanel[3].getTitleNumber()==0){
listenerPanel[0].setBackground(Color.white);
listenerPanel[0].setTitleNumber(0);
listenerPanel[3].setTitleNumber(m);
listenerPanel[3].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[1])){
if(listenerPanel[0].getTitleNumber()==0){
listenerPanel[1].setBackground(Color.white);
listenerPanel[1].setTitleNumber(0);
listenerPanel[0].setTitleNumber(m);
listenerPanel[0].setBackground(Color.darkGray);
}
if(listenerPanel[2].getTitleNumber()==0){
listenerPanel[1].setBackground(Color.white);
listenerPanel[1].setTitleNumber(0);
listenerPanel[2].setTitleNumber(m);
listenerPanel[2].setBackground(Color.darkGray);
}
if(listenerPanel[4].getTitleNumber()==0){
listenerPanel[1].setBackground(Color.white);
listenerPanel[1].setTitleNumber(0);
listenerPanel[4].setTitleNumber(m);
listenerPanel[4].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[2])){
if(listenerPanel[1].getTitleNumber()==0){
listenerPanel[2].setBackground(Color.white);
listenerPanel[2].setTitleNumber(0);
listenerPanel[1].setTitleNumber(m);
listenerPanel[1].setBackground(Color.darkGray);
}
if(listenerPanel[5].getTitleNumber()==0){
listenerPanel[2].setBackground(Color.white);
listenerPanel[2].setTitleNumber(0);
listenerPanel[5].setTitleNumber(m);
listenerPanel[5].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[3])){
if(listenerPanel[0].getTitleNumber()==0){
listenerPanel[3].setBackground(Color.white);
listenerPanel[3].setTitleNumber(0);
listenerPanel[0].setTitleNumber(m);
listenerPanel[0].setBackground(Color.darkGray);
}
if(listenerPanel[4].getTitleNumber()==0){
listenerPanel[3].setBackground(Color.white);
listenerPanel[3].setTitleNumber(0);
listenerPanel[4].setTitleNumber(m);
listenerPanel[4].setBackground(Color.darkGray);
}
if(listenerPanel[6].getTitleNumber()==0){
listenerPanel[3].setBackground(Color.white);
listenerPanel[3].setTitleNumber(0);
listenerPanel[6].setTitleNumber(m);
listenerPanel[6].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[4])){
if(listenerPanel[1].getTitleNumber()==0){
listenerPanel[4].setBackground(Color.white);
listenerPanel[4].setTitleNumber(0);
listenerPanel[1].setTitleNumber(m);
listenerPanel[1].setBackground(Color.darkGray);
}
if(listenerPanel[7].getTitleNumber()==0){
listenerPanel[4].setBackground(Color.white);
listenerPanel[4].setTitleNumber(0);
listenerPanel[7].setTitleNumber(m);
listenerPanel[7].setBackground(Color.darkGray);
}
if(listenerPanel[3].getTitleNumber()==0){
listenerPanel[4].setBackground(Color.white);
listenerPanel[4].setTitleNumber(0);
listenerPanel[3].setTitleNumber(m);
listenerPanel[3].setBackground(Color.darkGray);
}
if(listenerPanel[5].getTitleNumber()==0){
listenerPanel[4].setBackground(Color.white);
listenerPanel[4].setTitleNumber(0);
listenerPanel[5].setTitleNumber(m);
listenerPanel[5].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[5])){
if(listenerPanel[4].getTitleNumber()==0){
listenerPanel[5].setBackground(Color.white);
listenerPanel[5].setTitleNumber(0);
listenerPanel[4].setTitleNumber(m);
listenerPanel[4].setBackground(Color.darkGray);
}
if(listenerPanel[2].getTitleNumber()==0){
listenerPanel[5].setBackground(Color.white);
listenerPanel[5].setTitleNumber(0);
listenerPanel[2].setTitleNumber(m);
listenerPanel[2].setBackground(Color.darkGray);
}
if(listenerPanel[8].getTitleNumber()==0){
listenerPanel[5].setBackground(Color.white);
listenerPanel[5].setTitleNumber(0);
listenerPanel[8].setTitleNumber(m);
listenerPanel[8].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[6])){
if(listenerPanel[3].getTitleNumber()==0){
listenerPanel[6].setBackground(Color.white);
listenerPanel[6].setTitleNumber(0);
listenerPanel[3].setTitleNumber(m);
listenerPanel[3].setBackground(Color.darkGray);
}
if(listenerPanel[7].getTitleNumber()==0){
listenerPanel[6].setBackground(Color.white);
listenerPanel[6].setTitleNumber(0);
listenerPanel[7].setTitleNumber(m);
listenerPanel[7].setBackground(Color.darkGray);
}
}else if(jb.equals(listenerPanel[7])){
if(listenerPanel[6].getTitleNumber()==0){
listenerPanel[7].setBackground(Color.white);
listenerPanel[7].setTitleNumber(0);
listenerPanel[6].setTitleNumber(m);
listenerPanel[6].setBackground(Color.darkGray);
}
if(listenerPanel[8].getTitleNumber()==0){
listenerPanel[7].setBackground(Color.white);
listenerPanel[7].setTitleNumber(0);
listenerPanel[8].setTitleNumber(m);
listenerPanel[8].setBackground(Color.darkGray);
}
if(listenerPanel[4].getTitleNumber()==0){
listenerPanel[7].setBackground(Color.white);
listenerPanel[7].setTitleNumber(0);
listenerPanel[4].setTitleNumber(m);
listenerPanel[4].setBackground(Color.darkGray);
}
}else {
if(listenerPanel[5].getTitleNumber()==0){
listenerPanel[8].setBackground(Color.white);
listenerPanel[8].setTitleNumber(0);
listenerPanel[5].setTitleNumber(m);
listenerPanel[5].setBackground(Color.darkGray);
}
if(listenerPanel[7].getTitleNumber()==0){
listenerPanel[8].setBackground(Color.white);
listenerPanel[8].setTitleNumber(0);
listenerPanel[7].setTitleNumber(m);
listenerPanel[7].setBackground(Color.darkGray);
}
}
boolean b=true;//是否完成标记
for(int i=0;ilistenerPanel.length;i++){//判断listenerPanel[0]~listenerPanel[8]上的数字是从0~8.若是完成拼图
if(listenerPanel[i].getTitleNumber()!=i)
b=false;
}
if(b==true){
int i=JOptionPane.showConfirmDialog(null, "would you paly agin?");
if(i==0){
if(i==0){
Rectangle re=this.getBounds();
this.dispose();
SlidingPuzzle slidingPuzzle=new SlidingPuzzle();
slidingPuzzle.setBounds(re);
}
else if(i==1)
System.exit(0);
else ;
}
}
}
public void mouseReleased(MouseEvent e){}
public void mouseClicked(MouseEvent e){}
public void mouseEntered(MouseEvent e){}
public void mouseExited(MouseEvent e){}
}
如果运行过程什么问题追问或者hi
Q2: java的md5的加密算法代码
import java.lang.reflect.*;
/*******************************************************************************
* keyBean 类实现了RSA Data Security, Inc.在提交给IETF 的RFC1321中的keyBean message-digest
* 算法。
******************************************************************************/
public class keyBean {
/*
* 下面这些S11-S44实际上是一个4*4的矩阵，在原始的C实现中是用#define 实现的， 这里把它们实现成为static
* final是表示了只读，切能在同一个进程空间内的多个 Instance间共享
*/
static final int S11 = 7;
static final int S12 = 12;
static final int S13 = 17;
static final int S14 = 22;
static final int S21 = 5;
static final int S22 = 9;
static final int S23 = 14;
static final int S24 = 20;
static final int S31 = 4;
static final int S32 = 11;
static final int S33 = 16;
static final int S34 = 23;
static final int S41 = 6;
static final int S42 = 10;
static final int S43 = 15;
static final int S44 = 21;
static final byte[] PADDING = { -128, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0 };
/*
* 下面的三个成员是keyBean计算过程中用到的3个核心数据，在原始的C实现中 被定义到keyBean_CTX结构中
*/
private long[] state = new long[4]; // state (ABCD)
private long[] count = new long[2]; // number of bits, modulo 2^64 (lsb
// first)
private byte[] buffer = new byte[64]; // input buffer
/*
* digestHexStr是keyBean的唯一一个公共成员，是最新一次计算结果的 16进制ASCII表示.
*/
public String digestHexStr;
/*
* digest,是最新一次计算结果的2进制内部表示，表示128bit的keyBean值.
*/
private byte[] digest = new byte[16];
/*
* getkeyBeanofStr是类keyBean最主要的公共方法，入口参数是你想要进行keyBean变换的字符串
* 返回的是变换完的结果，这个结果是从公共成员digestHexStr取得的．
*/
public String getkeyBeanofStr(String inbuf) {
keyBeanInit();
keyBeanUpdate(inbuf.getBytes(), inbuf.length());
keyBeanFinal();
digestHexStr = "";
for (int i = 0; i 16; i++) {
digestHexStr += byteHEX(digest[i]);
}
return digestHexStr;
}
// 这是keyBean这个类的标准构造函数，JavaBean要求有一个public的并且没有参数的构造函数
public keyBean() {
keyBeanInit();
return;
}
/* keyBeanInit是一个初始化函数，初始化核心变量，装入标准的幻数 */
private void keyBeanInit() {
count[0] = 0L;
count[1] = 0L;
// /* Load magic initialization constants.
state[0] = 0x67452301L;
state[1] = 0xefcdab89L;
state[2] = 0x98badcfeL;
state[3] = 0x10325476L;
return;
}
/*
* F, G, H ,I 是4个基本的keyBean函数，在原始的keyBean的C实现中，由于它们是
* 简单的位运算，可能出于效率的考虑把它们实现成了宏，在java中，我们把它们 实现成了private方法，名字保持了原来C中的。
*/
private long F(long x, long y, long z) {
return (x y) | ((~x) z);
}
private long G(long x, long y, long z) {
return (x z) | (y (~z));
}
private long H(long x, long y, long z) {
return x ^ y ^ z;
}
private long I(long x, long y, long z) {
return y ^ (x | (~z));
}
/*
* FF,GG,HH和II将调用F,G,H,I进行近一步变换 FF, GG, HH, and II transformations for
* rounds 1, 2, 3, and 4. Rotation is separate from addition to prevent
* recomputation.
*/
private long FF(long a, long b, long c, long d, long x, long s, long ac) {
a += F(b, c, d) + x + ac;
a = ((int) a s) | ((int) a (32 - s));
a += b;
return a;
}
private long GG(long a, long b, long c, long d, long x, long s, long ac) {
a += G(b, c, d) + x + ac;
a = ((int) a s) | ((int) a (32 - s));
a += b;
return a;
}
private long HH(long a, long b, long c, long d, long x, long s, long ac) {
a += H(b, c, d) + x + ac;
a = ((int) a s) | ((int) a (32 - s));
a += b;
return a;
}
private long II(long a, long b, long c, long d, long x, long s, long ac) {
a += I(b, c, d) + x + ac;
a = ((int) a s) | ((int) a (32 - s));
a += b;
return a;
}
/*
* keyBeanUpdate是keyBean的主计算过程，inbuf是要变换的字节串，inputlen是长度，这个
* 函数由getkeyBeanofStr调用，调用之前需要调用keyBeaninit，因此把它设计成private的
*/
private void keyBeanUpdate(byte[] inbuf, int inputLen) {
int i, index, partLen;
byte[] block = new byte[64];
index = (int) (count[0] 3) 0x3F;
// /* Update number of bits */
if ((count[0] += (inputLen 3)) (inputLen 3))
count[1]++;
count[1] += (inputLen 29);
partLen = 64 - index;
// Transform as many times as possible.
if (inputLen = partLen) {
keyBeanMemcpy(buffer, inbuf, index, 0, partLen);
keyBeanTransform(buffer);
for (i = partLen; i + 63 inputLen; i += 64) {
keyBeanMemcpy(block, inbuf, 0, i, 64);
keyBeanTransform(block);
}
index = 0;
} else
i = 0;
// /* Buffer remaining input */
keyBeanMemcpy(buffer, inbuf, index, i, inputLen - i);
}
/*
* keyBeanFinal整理和填写输出结果
*/
private void keyBeanFinal() {
byte[] bits = new byte[8];
int index, padLen;
// /* Save number of bits */
Encode(bits, count, 8);
// /* Pad out to 56 mod 64.
index = (int) (count[0] 3) 0x3f;
padLen = (index 56) ? (56 - index) : (120 - index);
keyBeanUpdate(PADDING, padLen);
// /* Append length (before padding) */
keyBeanUpdate(bits, 8);
// /* Store state in digest */
Encode(digest, state, 16);
}
/*
* keyBeanMemcpy是一个内部使用的byte数组的块拷贝函数，从input的inpos开始把len长度的
* 字节拷贝到output的outpos位置开始
*/
private void keyBeanMemcpy(byte[] output, byte[] input, int outpos,
int inpos, int len) {
int i;
for (i = 0; i len; i++)
output[outpos + i] = input[inpos + i];
}
/*
* keyBeanTransform是keyBean核心变换程序，有keyBeanUpdate调用，block是分块的原始字节
*/
private void keyBeanTransform(byte block[]) {
long a = state[0], b = state[1], c = state[2], d = state[3];
long[] x = new long[16];
Decode(x, block, 64);
/* Round 1 */
a = FF(a, b, c, d, x[0], S11, 0xd76aa478L); /* 1 */
d = FF(d, a, b, c, x[1], S12, 0xe8c7b756L); /* 2 */
c = FF(c, d, a, b, x[2], S13, 0x242070dbL); /* 3 */
b = FF(b, c, d, a, x[3], S14, 0xc1bdceeeL); /* 4 */
a = FF(a, b, c, d, x[4], S11, 0xf57c0fafL); /* 5 */
d = FF(d, a, b, c, x[5], S12, 0x4787c62aL); /* 6 */
c = FF(c, d, a, b, x[6], S13, 0xa8304613L); /* 7 */
b = FF(b, c, d, a, x[7], S14, 0xfd469501L); /* 8 */
a = FF(a, b, c, d, x[8], S11, 0x698098d8L); /* 9 */
d = FF(d, a, b, c, x[9], S12, 0x8b44f7afL); /* 10 */
c = FF(c, d, a, b, x[10], S13, 0xffff5bb1L); /* 11 */
b = FF(b, c, d, a, x[11], S14, 0x895cd7beL); /* 12 */
a = FF(a, b, c, d, x[12], S11, 0x6b901122L); /* 13 */
d = FF(d, a, b, c, x[13], S12, 0xfd987193L); /* 14 */
c = FF(c, d, a, b, x[14], S13, 0xa679438eL); /* 15 */
b = FF(b, c, d, a, x[15], S14, 0x49b40821L); /* 16 */
/* Round 2 */
a = GG(a, b, c, d, x[1], S21, 0xf61e2562L); /* 17 */
d = GG(d, a, b, c, x[6], S22, 0xc040b340L); /* 18 */
c = GG(c, d, a, b, x[11], S23, 0x265e5a51L); /* 19 */
b = GG(b, c, d, a, x[0], S24, 0xe9b6c7aaL); /* 20 */
a = GG(a, b, c, d, x[5], S21, 0xd62f105dL); /* 21 */
d = GG(d, a, b, c, x[10], S22, 0x2441453L); /* 22 */
c = GG(c, d, a, b, x[15], S23, 0xd8a1e681L); /* 23 */
b = GG(b, c, d, a, x[4], S24, 0xe7d3fbc8L); /* 24 */
a = GG(a, b, c, d, x[9], S21, 0x21e1cde6L); /* 25 */
d = GG(d, a, b, c, x[14], S22, 0xc33707d6L); /* 26 */
c = GG(c, d, a, b, x[3], S23, 0xf4d50d87L); /* 27 */
b = GG(b, c, d, a, x[8], S24, 0x455a14edL); /* 28 */
a = GG(a, b, c, d, x[13], S21, 0xa9e3e905L); /* 29 */
d = GG(d, a, b, c, x[2], S22, 0xfcefa3f8L); /* 30 */
c = GG(c, d, a, b, x[7], S23, 0x676f02d9L); /* 31 */
b = GG(b, c, d, a, x[12], S24, 0x8d2a4c8aL); /* 32 */
/* Round 3 */
a = HH(a, b, c, d, x[5], S31, 0xfffa3942L); /* 33 */
d = HH(d, a, b, c, x[8], S32, 0x8771f681L); /* 34 */
c = HH(c, d, a, b, x[11], S33, 0x6d9d6122L); /* 35 */
b = HH(b, c, d, a, x[14], S34, 0xfde5380cL); /* 36 */
a = HH(a, b, c, d, x[1], S31, 0xa4beea44L); /* 37 */
d = HH(d, a, b, c, x[4], S32, 0x4bdecfa9L); /* 38 */
c = HH(c, d, a, b, x[7], S33, 0xf6bb4b60L); /* 39 */
b = HH(b, c, d, a, x[10], S34, 0xbebfbc70L); /* 40 */
a = HH(a, b, c, d, x[13], S31, 0x289b7ec6L); /* 41 */
d = HH(d, a, b, c, x[0], S32, 0xeaa127faL); /* 42 */
c = HH(c, d, a, b, x[3], S33, 0xd4ef3085L); /* 43 */
b = HH(b, c, d, a, x[6], S34, 0x4881d05L); /* 44 */
a = HH(a, b, c, d, x[9], S31, 0xd9d4d039L); /* 45 */
d = HH(d, a, b, c, x[12], S32, 0xe6db99e5L); /* 46 */
c = HH(c, d, a, b, x[15], S33, 0x1fa27cf8L); /* 47 */
b = HH(b, c, d, a, x[2], S34, 0xc4ac5665L); /* 48 */
/* Round 4 */
a = II(a, b, c, d, x[0], S41, 0xf4292244L); /* 49 */
d = II(d, a, b, c, x[7], S42, 0x432aff97L); /* 50 */
c = II(c, d, a, b, x[14], S43, 0xab9423a7L); /* 51 */
b = II(b, c, d, a, x[5], S44, 0xfc93a039L); /* 52 */
a = II(a, b, c, d, x[12], S41, 0x655b59c3L); /* 53 */
d = II(d, a, b, c, x[3], S42, 0x8f0ccc92L); /* 54 */
c = II(c, d, a, b, x[10], S43, 0xffeff47dL); /* 55 */
b = II(b, c, d, a, x[1], S44, 0x85845dd1L); /* 56 */
a = II(a, b, c, d, x[8], S41, 0x6fa87e4fL); /* 57 */
d = II(d, a, b, c, x[15], S42, 0xfe2ce6e0L); /* 58 */
c = II(c, d, a, b, x[6], S43, 0xa3014314L); /* 59 */
b = II(b, c, d, a, x[13], S44, 0x4e0811a1L); /* 60 */
a = II(a, b, c, d, x[4], S41, 0xf7537e82L); /* 61 */
d = II(d, a, b, c, x[11], S42, 0xbd3af235L); /* 62 */
c = II(c, d, a, b, x[2], S43, 0x2ad7d2bbL); /* 63 */
b = II(b, c, d, a, x[9], S44, 0xeb86d391L); /* 64 */
state[0] += a;
state[1] += b;
state[2] += c;
state[3] += d;
}
/*
* Encode把long数组按顺序拆成byte数组，因为java的long类型是64bit的， 只拆低32bit，以适应原始C实现的用途
*/
private void Encode(byte[] output, long[] input, int len) {
int i, j;
for (i = 0, j = 0; j len; i++, j += 4) {
output[j] = (byte) (input[i] 0xffL);
output[j + 1] = (byte) ((input[i] 8) 0xffL);
output[j + 2] = (byte) ((input[i] 16) 0xffL);
output[j + 3] = (byte) ((input[i] 24) 0xffL);
}
}
/*
* Decode把byte数组按顺序合成成long数组，因为java的long类型是64bit的，
* 只合成低32bit，高32bit清零，以适应原始C实现的用途
*/
private void Decode(long[] output, byte[] input, int len) {
int i, j;
for (i = 0, j = 0; j len; i++, j += 4)
output[i] = b2iu(input[j]) | (b2iu(input[j + 1]) 8)
| (b2iu(input[j + 2]) 16) | (b2iu(input[j + 3]) 24);
return;
}
/*
* b2iu是我写的一个把byte按照不考虑正负号的原则的”升位”程序，因为java没有unsigned运算
*/
public static long b2iu(byte b) {
return b 0 ? b 0x7F + 128 : b;
}
/*
* byteHEX()，用来把一个byte类型的数转换成十六进制的ASCII表示，
* 因为java中的byte的toString无法实现这一点，我们又没有C语言中的 sprintf(outbuf,"%02X",ib)
*/
public static String byteHEX(byte ib) {
char[] Digit = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A',
'B', 'C', 'D', 'E', 'F' };
char[] ob = new char[2];
ob[0] = Digit[(ib 4) 0X0F];
ob[1] = Digit[ib 0X0F];
String s = new String(ob);
return s;
}
public static void main(String args[]) {
keyBean m = new keyBean();
if (Array.getLength(args) == 0) { // 如果没有参数，执行标准的Test Suite
System.out.println("keyBean Test suite:");
System.out.println("keyBean(\"):" + m.getkeyBeanofStr(""));
System.out.println("keyBean(\"a\"):" + m.getkeyBeanofStr("a"));
System.out.println("keyBean(\"abc\"):" + m.getkeyBeanofStr("abc"));
System.out.println("keyBean(\"message digest\"):"
+ m.getkeyBeanofStr("message digest"));
System.out.println("keyBean(\"abcdefghijklmnopqrstuvwxyz\"):"
+ m.getkeyBeanofStr("abcdefghijklmnopqrstuvwxyz"));
System.out
.println("keyBean(\"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789\"):"
+ m
.getkeyBeanofStr("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789"));
} else
System.out.println("keyBean(" + args[0] + ")="
+ m.getkeyBeanofStr(args[0]));
}
}
Q3: 用java实现哈夫曼编码
只要自己再加个类Tree就可以了。
代码如下：
public class Tree {
double lChild, rChild, parent;
public Tree (double lChild, double rChild, double parent) {
this.lChild = lChild;
this.rChild = rChild;
this.parent = parent;
}
public double getLchild() {
return lChild;
}
public void setLchild(double lChild) {
this.lChild = lChild;
}
public double getRchild() {
return rChild;
}
public void setRchild(double rChild) {
this.rChild = rChild;
}
public double getParents() {
return parent;
}
public void setParents(double root) {
this.parent = root;
}
}
Q4: 用java求最短路径问题，求源程序
import java.util.Vector;
public class Link {
private Vector link = new Vector();
// private Link next = null;
public Link() {
}
public boolean addNode(Node setNode){//增加一个节点
setNode = checkNode(setNode);
if(setNode != null){
this.link.addElement((Node)setNode);
return true;
}
return false;
}
public void delNode(Node setNode){ //删除一个节点
if(!this.link.isEmpty()){
for(int i=0;i this.link.size(); i++)
{
if(setNode.getPos() == ((Node)this.link.elementAt(i)).getPos()){
this.link.remove(i);
//System.out.println("asdfasdfas:"+this.link.size());
break;
}
}
}
}
public Node checkNode(Node setNode){//判断节点是否在链表里面并取得两者的最佳值
if(!this.link.isEmpty() setNode!=null){
for(int i=0;i this.link.size(); i++)
{
if(setNode.getPos() == ((Node)this.link.elementAt(i)).getPos()){
if(setNode.getStep() ((Node)this.link.elementAt(i)).getStep()){
setNode = (Node)this.link.elementAt(i);
this.link.remove(i);
}
else
return null;
break;
}
}
}
return setNode;
}
public boolean isEmpty(){
return this.link.isEmpty();
}
public Node getBestNode(){ //得到最好的节点
Node tmpNode = null;
if(!this.link.isEmpty()){
tmpNode = (Node)this.link.elementAt(0);
//System.out.println("tmpNodeStep:"+tmpNode.getStep());
//System.out.print("OpenNode(pos,step):");
for(int i=1;i this.link.size(); i++)
{
//System.out.print("("+((Node)this.link.elementAt(i)).getPos()+","+((Node)this.link.elementAt(i)).getStep()+")");
if(tmpNode.getJudgeNum() = ((Node)this.link.elementAt(i)).getJudgeNum()){
tmpNode = (Node)this.link.elementAt(i);
}
}
}
return tmpNode;
}
}
public class FindBestPath {
private char[][] map = null;//地图
private int maxX,maxY;//最大的地图边界大小
Node startNode = null;//入口
Node endNode = null;//出口
private int endX,endY;
/*初始化
*@param setMap 地图
*@param setX,setY 边界值
//////////*@param startNode 入口
//////////*param endNode 出口
*@param sX,sY:开始点
*@param eX,eY:结束点
*/
public FindBestPath(char[][] setMap,int setX,int setY,int sX,int sY,int eX,int eY) {
this.map = setMap;
this.maxY = setX - 1; //x,y互换
this.maxX = setY - 1; //x,y互换
//this.startNode = sNode;
//this.endNode = eNode;
Node sNode = new Node();
Node eNode = new Node();
sNode.setFarther(null);
sNode.setPos(posToNum(sX,sY));
sNode.setStep(0);
eNode.setPos(posToNum(eX,eY));
this.startNode = sNode;
this.endNode = eNode;
this.endX = eX;//numToX(eNode.getPos());
this.endY = eY;//numToY(eNode.getPos());
}
public int posToNum(int x,int y){//从xy坐标获得编号
return (x+y*(this.maxY+1));
}
public int numToX(int num){//从编号获得x坐标
return (num%(this.maxY+1));
}
public int numToY(int num){//从编号获得y坐标
return (int)(num/(this.maxY+1));
}
public boolean checkVal(int x,int y){//判断是否为障碍
//System.out.println("map["+x+"]["+y+"]="+map[x][y]);
if(this.map[x][y] == 'N')
return false;
else
return true;
}
public int judge(Node nowNode){//一定要比实际距离小
//System.out.println("nowNodePos:"+nowNode.getPos());
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
int distance = Math.abs((nowX-this.endX))+Math.abs((nowY-this.endY));
// System.out.println("distance:"+distance);
return distance;
}
public Node getLeft(Node nowNode){//取得左节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowY 0){//判断节点是否到最左
if(checkVal(nowX,nowY-1)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX,nowY-1));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Node getRight(Node nowNode){//取得右节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowY this.maxX){//判断节点是否到最左
if(checkVal(nowX,nowY+1)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX,nowY+1));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Node getTop(Node nowNode){//取得上节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowX 0){//判断节点是否到最左
if(checkVal(nowX-1,nowY)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX-1,nowY));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Node getBottom(Node nowNode){//取得下节点
int nowX = numToX(nowNode.getPos());
int nowY = numToY(nowNode.getPos());
Node tmpNode = new Node();
if(nowX this.maxY){//判断节点是否到最左
if(checkVal(nowX+1,nowY)){
tmpNode.setFarther(nowNode);
tmpNode.setPos(posToNum(nowX+1,nowY));
tmpNode.setStep(nowNode.getStep()+1);
tmpNode.setJudgeNum(tmpNode.getStep()+judge(tmpNode));
return tmpNode;
}
}
return null;
}
public Link getBestPath(){//寻找路径
Link openLink = new Link();//没有访问的路径
Link closeLink = new Link();//访问过的路径
Link path = null;//最短路径
Node bestNode = null;
Node tmpNode = null;
openLink.addNode(this.startNode);
while(!openLink.isEmpty())//openLink is not null
{
bestNode = openLink.getBestNode();//取得最好的节点
//System.out.println("bestNode:("+numToX(bestNode.getPos())+","+numToY(bestNode.getPos())+")step:"+bestNode.getJudgeNum());
if(bestNode.getPos()==this.endNode.getPos())
{
/*this.endNode.setStep(bestNode.getStep()+1);
this.endNode.setFarther(bestNode);
this.endNode.setJudgeNum(bestNode.getStep()+1);*/
path = makePath(bestNode);
break;
}
else
{
tmpNode = closeLink.checkNode(getLeft(bestNode));
if(tmpNode != null)
//System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
tmpNode = closeLink.checkNode(getRight(bestNode));
if(tmpNode != null)
// System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
tmpNode = closeLink.checkNode(getTop(bestNode));
if(tmpNode != null)
// System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
tmpNode = closeLink.checkNode(getBottom(bestNode));
if(tmpNode != null)
// System.out.println("("+numToY(tmpNode.getPos())+","+numToX(tmpNode.getPos())+")");
openLink.addNode(tmpNode);
openLink.delNode(bestNode);
closeLink.addNode(bestNode);
}
}
return path;
}
public Link makePath(Node lastNode){//制造路径
Link tmpLink = new Link();
Node tmpNode = new Node();
int x,y;
tmpNode = lastNode;
if(tmpNode != null){
do{
x=numToX(tmpNode.getPos());
y=numToY(tmpNode.getPos());
System.out.println("map["+x+"]["+y+"]="+map[x][y]);
tmpLink.addNode(tmpNode);
tmpNode = tmpNode.getFarther();
}while(tmpNode != null);
}else
{
System.out.println("Couldn't find the path!");
}
return tmpLink;
}
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
char[][] map ={
{'Y', 'N', 'z', 'y', 'x', 'w', 'v', 'N', 'N', 'N'},
{'Y', 'N', '1', 'N', 'N', 'N', 'u', 't', 'N', 'N'},
{'N', '1', '2', '1', '1', '1', 'N', 's', 'N', 'N'},
{'N', 'N', '1', 'N', '9', 'N', 'q', 'r', 'N', 'N'},
{'N', 'N', '1', 'N', 'n', 'o', 'p', 'N', 'N', 'N'},
{'N', '4', '5', '6', 'm', 'N', 'N', 'N', 'N', 'N'},
{'N', '3', 'N', '5', 'l', 'k', 'j', 'N', 'N', 'N'},
{'N', 'N', '3', '4', 'N', 'd', 'i', 'd', 'N', 'N'},
{'N', '1', 'N', 'N', '1', 'N', 'h', 'N', 'N', 'N'},
{'N', '1', 'N', 'N', '1', 'N', 'g', 'N', 'N', 'N'},
{'N', 'a', 'b', 'c', 'd', 'e', 'f', 'N', 'N', 'N'}
};
/*map[x][y]
*如上所示：maxY=10 maxX=11 横的代表maxY,竖的代表maxX 可以自己替换
*地图的读取是
*for(i=1;i行的最大值;i++)
* for(j=1;j列的最大值;j++)
* map[i][j] = 地图[i][j]
*/
Link bestPath = new Link();
/*startNode.setFarther(null);
startNode.setPos(21);
startNode.setStep(0);
//endNode.setFarther(startNode);
endNode.setPos(79);
//endNode.setStep(0);*/
FindBestPath path = new FindBestPath(map, 11, 10, 10, 1, 0, 2);
//FindBestPath path = new FindBestPath(map, 11, 10, startNode, endNode);
bestPath = path.getBestPath();
//bestPath.printLink();
}
}
public class Node {
private int step;//从入口到该节点经历的步数
private int pos;//位置
private Node farther;//上一个结点
private int judgeNum;
public Node() {
}
public void setStep(int setStep){
this.step = setStep;
}
public int getStep(){
return this.step;
}
public void setPos(int setPos){
this.pos = setPos;
}
public int getPos(){
return this.pos;
}
public void setFarther(Node setNode){
this.farther = setNode;;
}
public Node getFarther(){
return this.farther;
}
public void setJudgeNum (int setInt){
this.judgeNum = setInt;;
}
public int getJudgeNum(){
return this.judgeNum;
}
}
Q5: 匈牙利算法 java
#includestdio.h
#includestring.h
bool g[201][201];
int n,m,ans;
bool b[201];
int link[201];
bool init()
{
int _x,_y;
memset(g,0,sizeof(g));
memset(link,0,sizeof(link));
ans=0;
if(scanf("%d%d",n,m)==EOF)return false;
for(int i=1;i=n;i++)
{
scanf("%d",_x);
for(int j=0;j_x;j++)
{
scanf("%d",_y);
g[ i ][_y]=true;
}
}
return true;
}
bool find(int a)
{
for(int i=1;i=m;i++)
{
if(g[a][ i ]==1!b[ i ])
{
b[ i ]=true;
if(link[ i ]==0||find(link[ i ]))
{
link[ i ]=a;
return true;
}
}
}
return false;
}
int main()
{
while(init())
{
for(int i=1;i=n;i++)
{
memset(b,0,sizeof(b));
if(find(i))ans++;
}
printf("%d\n",ans);
}
}
Pascal:
Program matching;
Const
max = 1000;
Var
map : array [1..max, 1..max] of boolean; {邻接矩阵}
match: array [1..max] of integer; {记录当前连接方式}
chk : array [1..max] of boolean; {记录是否遍历过，防止死循环}
m, n, i, t1, t2, ans,k: integer;
Function dfs(p: integer): boolean;
var
i, t: integer;
begin
for i:=1 to k do
if map[p, i] and not chk[ i ] then
begin
chk[ i ] := true;
if (match[ i ] = 0) or dfs(match[ i ]) then {没有被连过 或 寻找到增广路}
begin
match[ i ] := p;
exit(true);
end;{if}
end;{for}
exit(false);
end;{function}
begin{main}
readln(n, m); {N 为二分图左侧点数 M为可连接的边总数}
fillchar(map, sizeof(map), 0);
k:=0;
for i:=1 to m do{initalize}
begin
readln(t1, t2);
map[t1, t2] := true;
if kt2 then k:=t2;
end;{for}
fillchar(match, sizeof(match), 0);
ans := 0;
for i:=1 to n do
begin
fillchar(chk, sizeof(chk), 0);
if dfs(i) then inc(ans);
end;
writeln(ans);
for i:=1 to 1000 do
if match[ i ] 0 then
writeln(match[ i ], '--', i);
end.
