哈夫曼编码译码

课程设计：

已知信源的概率分布如下

构造其二元的Huffman编码器和译码器。

以下是源程序，在VC6.0编译器下通过。

# include <stdio.h># include <string.h># include <iostream.h># define N 50         //哈夫曼树叶子结点数# define M 2*N-1      //树中结点数typedef struct{char data[5];double weight;int parent;int lchild;int rchild;}HTNode;typedef struct{char cd[N];          //存放哈夫曼码,N值为最大码长,可在定义宏中更改int start;}HCode;void CreateHT(HTNode ht[],int n)   //构造哈夫曼树{int i,k,lnode,rnode;double min1,min2;for(i=0;i<2*n-1;i++)ht[i].parent=ht[i].lchild=ht[i].rchild=-1;for(i=n;i<2*n-1;i++){min1=min2=1.0;lnode=rnode=-1;       //lnode和rnode为最小权重的两个结点位置for(k=0;k<i;k++)              //在还没构造二叉树的结点中查找if(ht[k].parent==-1){if(ht[k].weight<min1){min2=min1;rnode=lnode;min1=ht[k].weight;lnode=k;}else if(ht[k].weight<min2){min2=ht[k].weight;rnode=k;}}ht[lnode].parent=i;ht[rnode].parent=i;ht[i].weight=ht[lnode].weight+ht[rnode].weight;ht[i].lchild=lnode;ht[i].rchild=rnode;}}void CreateHCode(HTNode ht[],HCode hcd[],int n)     //构造哈夫曼编码{int i,f,c;HCode hc;for(i=0;i<n;i++){hc.start=n;c=i;f=ht[i].parent;while(f!=-1){if(ht[f].lchild==c)                      //处理左孩子hc.cd[hc.start--]='1';else                                     //处理右孩子hc.cd[hc.start--]='0';c=f;f=ht[f].parent;}hc.start++;hcd[i]=hc;}}void TransHCode(HTNode ht[],HCode hcd[], int n)    //译码{     int m=2*n-1;     int i,j=0,k; char c[50]={'0','1','0','1','1','0','1','1','1','1','1','0','0'};      //初始化输入码,可更改,为'0,'1'级组合 cout<<"/t假设要译的码为 : "; while(c[j]!='/0')     { cout<<c[j]; j++; } cout<<endl; j=0;       //重置为0 cout<<"/t译码结果为 : "; while(c[j]!='/0')     {         i=m-1;         while(ht[i].lchild!=-1&&ht[i].rchild!=-1) { if(c[j]=='0') i=ht[i].rchild; else if(c[j]=='1')                 i=ht[i].lchild;             j++; }         cout<<ht[i].data<<i+1<<" ";     }     cout<<endl<<endl<<endl;}void DispHCode(HTNode ht[],HCode hcd[],int n)           //编码输出{int i,j,k;cout<<"/t信源   "<<"概率   "<<"编码"<<endl;for(i=0;i<n;i++){j=0;cout<<"/t "<<ht[i].data<<i+1<<"    "<<ht[i].weight<<"   ";for(k=hcd[i].start;k<=n;k++){cout<<hcd[i].cd[k];j++;}cout<<endl;}/******************************************************************************此处代码供调试用,可删除********************for(i=0;i<2*n-1;i++){cout<<"ht["<<i<<"].lchild  "<<ht[i].lchild<<" ";cout<<"ht["<<i<<"].rchild  "<<ht[i].rchild<<" ";cout<<"ht["<<i<<"].parent  "<<ht[i].parent<<" ";cout<<"ht["<<i<<"].weight  "<<ht[i].weight<<" "<<endl;}*************************************************************/cout<<endl;}void interface()              //界面初始化{cout<<endl<<endl;cout<<"/t***********************************************************"<<endl;    cout<<"/t******                 哈夫曼编码译码              ********"<<endl;    cout<<"/t******    XXXXX大学   XXXXXXXXXX学院   XXXXXXX班   ********"<<endl;    cout<<"/t******             组成员： AAAA  BBBB CCCC        ********"<<endl;    cout<<"/t******                 指导老师：TTTT              ********"<<endl;    cout<<"/t******                   NNNN年N月N日              ********"<<endl;    cout<<"/t***********************************************************"<<endl;cout<<endl<<endl;}void main(){interface();               //调用函数导入界面/***************************初始化***********************************/int n=6;           //权值个数,此处也即信源概率个数,可根据情况更改//char *ch=new char[N];double fnum[]={0.32,0.22,0.18,0.16,0.08,0.04};    //权值(信源概率分布)HTNode ht[M];HCode hcd[N];for(int i=0;i<n;i++){strcpy(ht[i].data,"S");ht[i].weight=fnum[i];}    /*********************************************************************/CreateHT(ht,n);             //调用函数构造哈夫曼树CreateHCode(ht,hcd,n);      //调用函数构造哈夫曼编码DispHCode(ht,hcd,n);        //调用函数输出哈夫曼编码    TransHCode(ht,hcd,n);    //调用函数译码}