哈夫曼算法原理

Wikipedia上面说的非常清楚了，这里我就不再赘述，直接贴过来了。

1952年, David A. Huffman提出了一个不同的算法，这个算法能够为不论什么的可能性提供出一个理想的树。香农-范诺编码（Shanno-Fano）是从树的根节点到叶子节点所进行的的编码，哈夫曼编码算法却是从相反的方向，暨从叶子节点到根节点的方向编码的。

1. 为每一个符号建立一个叶子节点，并加上其对应的发生频率
2. 当有一个以上的节点存在时，进行下列循环:
   1. 把这些节点作为带权值的二叉树的根节点，左右子树为空
   2. 选择两棵根结点权值最小的树作为左右子树构造一棵新的二叉树，且至新的二叉树的根结点的权值为其左右子树上根结点的权值之和。
   3. 把权值最小的两个根节点移除
   4. 将新的二叉树增加队列中.
3. 最后剩下的节点暨为根节点，此时二叉树已经完毕。

演示样例

Huffman Algorithm

符号	A	B	C	D	E
计数	15	7	6	6	5
概率	0.38461538	0.17948718	0.15384615	0.15384615	0.12820513

在这样的情况下，D，E的最低频率和分配分别为0和1，分组结合概率的0.28205128。如今最低的一双是B和C，所以他们就分配0和1组合结合概率的0.33333333在一起。这使得BC和DE所以0和1的前面加上他们的代码和它们结合的概率最低。然后离开仅仅是一个和BCDE，当中有前缀分别为0和1，然后结合。这使我们与一个单一的节点，我们的算法是完整的。

可得A代码的代码长度是1比特，其余字符是3比特。

字符	A	B	C	D	E
代码	0	100	101	110	111

   

Entropy：

Pseudo-code

1:  begin 2:     count frequencies of single characters (source units) 3:     output(frequencies using Fibonacci Codes of degree 2) 4:     sort them to non-decreasing sequence 5:     create a leaf node (character, frequency c, left son = NULL, right son = NULL)  6:  	   of the tree for each character and put nodes into queue F 7:     while (|F|>=2) do  8:      begin 9:        pop the first two nodes (u1, u2) with the lowest 10:  	      frequencies from sorted queue11:        create a node evaluated with sum of the chosen units, 12:  	      successors are chosen units (eps, c(u1)+c(u2), u1, u2)13:        insert new node into queue14:      end15:     node evaluate with way from root to leaf node (left son 1, right son 0)16:     create output from coded intput characters17:  end

哈夫曼算法实现

实现的时候我们用vector<bool>记录每一个char的编码；用map<char,vector<bool>>表示整个字典；

就得到了以下的代码（以下有两个，一对一错）：

先放出来这个错的，以示警戒：

/************************************************************************//*	File Name: Huffman.cpp*		@Function: Lossless Compression		@Author: Sophia Zhang		@Create Time: 2012-9-26 10:40		@Last Modify: 2012-9-26 11:10*//************************************************************************/#include"iostream"#include "queue"#include "map"#include "string"#include "iterator"#include "vector"#include "algorithm"using namespace std;#define NChar 8	//suppose use at most 8 bits to describe all symbols#define Nsymbols 1<
      
        Huff_code;//8 bit code of one charmap
       
         Huff_Dic;	//huffman coding dictionaryclass HTree{public :	HTree* left;	HTree* right;	char ch;	int weight;	HTree(){left = right = NULL; weight=0;}	HTree(HTree* l,HTree* r,int w,char c){left = l;	right = r;	weight=w;	ch=c;}	~HTree(){delete left; delete right;}	int Getweight(){return weight?weight:left->weight+right->weight;}	bool Isleaf(){return !left && !right; }	bool operator < (const HTree tr) const	{		return tr.weight < weight;	}};HTree* BuildTree(int *frequency){	priority_queue
        
          QTree;	//1st level add characters	for (int i=0;i
         
          1)	{		HTree* lc  = QTree.top();		QTree.pop();		HTree* rc = QTree.top();		QTree.pop();		HTree* parent = new HTree(lc,rc,parent->Getweight(),(char)256);		QTree.push(parent);	}	//return tree root	return QTree.top();}void Huffman_Coding(HTree* root, Huff_code& curcode){	if(root->Isleaf())	{		Huff_Dic[root->ch] = curcode;		return;	}	Huff_code& lcode = curcode;	Huff_code& rcode = curcode;	lcode.push_back(false);	rcode.push_back(true);	Huffman_Coding(root->left,lcode);	Huffman_Coding(root->right,rcode);}int main(){	int freq[Nsymbols] = {0};	char *str = "this is the string need to be compressed";	//statistic character frequency	while (*str!='\0')		freq[*str++]++;	//build tree	HTree* r = BuildTree(freq);	Huff_code nullcode;	nullcode.clear();	Huffman_Coding(r,nullcode);	for(map
          
           ::iterator it = Huff_Dic.begin(); it != Huff_Dic.end(); it++)	{		cout<<(*it).first<<'\t';		Huff_code vec_code = (*it).second;		for (vector
           
            ::iterator vit = vec_code.begin(); vit!=vec_code.end();vit++) { cout<<(*vit)<
           
          
         
        
       
      

上面这段代码，我执行出来不正确。在调试的时候发现了一个问题，就是QTree优先队列中的排序出了问题，说来也是，上面的代码中，我重载小于号是对HTree object做的；而实际上我建树时用的是指针，那么优先级队列中元素为指针时该怎么办呢？

那我们将friend bool operator >(Node node1,Node node2)改动为friend bool operator >(Node* node1,Node* node2)，也就是传递的是Node的指针行不行呢?

答案是不能够，由于依据c++primer中重载操作符中讲的“程序猿仅仅能为类类型或枚举类型的操作数定义重载操作符，在把操作符声明为类的成员时，至少有一个类或枚举类型的參数依照值或者引用的方式传递”，也就是说friend bool operator >(Node* node1,Node* node2)形參中都是指针类型的是不能够的。我们仅仅能再建一个类，用当中的重载（）操作符作为优先队列的比較函数。

就得到了以下正确的代码：

/************************************************************************//*	File Name: Huffman.cpp*		@Function: Lossless Compression		@Author: Sophia Zhang		@Create Time: 2012-9-26 10:40		@Last Modify: 2012-9-26 12:10*//************************************************************************/#include"iostream"#include "queue"#include "map"#include "string"#include "iterator"#include "vector"#include "algorithm"using namespace std;#define NChar 8	//suppose use 8 bits to describe all symbols#define Nsymbols 1<
      
        Huff_code;//8 bit code of one charmap
       
         Huff_Dic;	//huffman coding dictionary/************************************************************************//* Tree Class elements:*2 child trees*character and frequency of current node*//************************************************************************/class HTree{public :	HTree* left;	HTree* right;	char ch;	int weight;	HTree(){left = right = NULL; weight=0;ch ='\0';}	HTree(HTree* l,HTree* r,int w,char c){left = l;	right = r;	weight=w;	ch=c;}	~HTree(){delete left; delete right;}	bool Isleaf(){return !left && !right; }};/************************************************************************//* prepare for pointer sorting*//*because we cannot use overloading in class HTree directly*//************************************************************************/class Compare_tree{public:	bool operator () (HTree* t1, HTree* t2)	{		return t1->weight> t2->weight;	}};/************************************************************************//* use priority queue to build huffman tree*//************************************************************************/HTree* BuildTree(int *frequency){	priority_queue
        
         
          ,Compare_tree> QTree;	//1st level add characters	for (int i=0;i
          
           1)	{		HTree* lc  = QTree.top();		QTree.pop();		HTree* rc = QTree.top();		QTree.pop();		HTree* parent = new HTree(lc,rc,lc->weight+rc->weight,(char)256);		QTree.push(parent);	}	//return tree root	return QTree.top();}/************************************************************************//* Give Huffman Coding to the Huffman Tree*//************************************************************************/void Huffman_Coding(HTree* root, Huff_code& curcode){	if(root->Isleaf())	{		Huff_Dic[root->ch] = curcode;		return;	}	Huff_code lcode = curcode;	Huff_code rcode = curcode;	lcode.push_back(false);	rcode.push_back(true);	Huffman_Coding(root->left,lcode);	Huffman_Coding(root->right,rcode);}int main(){	int freq[Nsymbols] = {0};	char *str = "this is the string need to be compressed";	//statistic character frequency	while (*str!='\0')		freq[*str++]++;	//build tree	HTree* r = BuildTree(freq);	Huff_code nullcode;	nullcode.clear();	Huffman_Coding(r,nullcode);	for(map
           
            ::iterator it = Huff_Dic.begin(); it != Huff_Dic.end(); it++) { cout<<(*it).first<<'\t'; std::copy(it->second.begin(),it->second.end(),std::ostream_iterator
            
             (cout)); cout<
            
           
          
         
        
       
      

Reference：

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