谁能个给下面的哈夫曼程序加一段译码的代码，求高手热心人

Question

#include

#include

#include

using namespace std;

#define N 10 // 带编码字符的个数，即树中叶结点的最大个数

#define M (2*N-1) // 树中总的结点数目

class HTNode{ // 树中结点的结构

public:
char data;

int weight;

int parent,lchild,rchild;

};

class HTCode{

public:

char data; // 待编码的字符

int weight; // 字符的权值

char code[N];
int start; // 字符的编码

};

void Init(HTCode hc[], int *n){

// 初始化，读入待编码字符的个数n，从键盘输入n个字符和n个权值

int i;

cout<<"输入字符个数：";
cin>>*n;

cout<<"输入"<<*n<<"个字符"; 

for(i=1; i<=*n; ++i)  
    cin>>hc[i].data;

cout<<"输入"<<*n<<"个权值";  

for(i=1; i<=*n; ++i)  
    cin>>hc[i].weight; 

}

void Select(HTNode ht[], int k, int *s1, int *s2){

// ht[1...k]中选择parent为0，并且weight最小的两个结点，其序号由指针变量s1，s2指示

int i;

for(i=1; i<=k && ht[i].parent != 0; i++){}

*s1 = i;

for(i=1; i<=k; ++i){  
    if(ht[i].parent==0 && ht[i].weight<ht[*s1].weight)  
        *s1 = i;  
}  

for(i=1; i<=k; i++){  
    if(ht[i].parent==0 && i!=*s1)  
        break;  
}  
*s2 = i;  

for(i=1; i<=k; i++){  
    if(ht[i].parent==0 && i!=*s1 && ht[i].weight<ht[*s2].weight)  
        *s2 = i;  
}  

}
void print(HTNode ht[],int k ,int n)
{
if(k==0)
{
return;
}
for (int i=0;i<n;i++)
{
cout<<" ";
}
if(ht[k].parent!=-1)
{
cout<<"|_";
}
else {
cout<<" ";
}
cout<<ht[k].data<<endl;
int L=ht[k].lchild;
int R=ht[k].rchild;
print(ht,R,n+2);
print(ht,L,n+2);
}

void HuffmanCoding(HTNode ht[],HTCode hc[],int n){

// 构造Huffman树ht，并求出n个字符的编码

char cd[N];
HTCode d;

int i,j,m,c,f,s1,s2,start;

for(i=1; i<=2*n-1; i++)
{  
    if(i <= n)  
        ht[i].weight = hc[i].weight;  
    else  
        ht[i].parent = 0;  
    ht[i].parent = ht[i].lchild = ht[i].rchild = 0;  
}
for(i=n+1; i<=2*n-1; i++){  
    Select(ht, i-1, &s1, &s2);  
    ht[s1].parent = i;  
    ht[s2].parent = i;  
    ht[i].lchild = s1;  
    ht[i].rchild = s2;  
    ht[i].weight = ht[s1].weight+ht[s2].weight;  
}  cd[n-1] = '\0';  

for(i=1; i<=n; i++){  
    start = n-1;  
    for(c=i,f=ht[i].parent; f; c=f,f=ht[f].parent){  
        if(ht[f].lchild == c)  
            cd[--start] = '0';  
        else  
            cd[--start] = '1';  
    }  
    strcpy(hc[i].code, &cd[start]);  
}
//dayin(ht, 2*n-1,0 );   

}

int main(int i, int n )
{

int m,w[N+1];

HTNode ht[M+1];

HTCode hc[N+1];

Init(hc, &n); // 初始化

HuffmanCoding(ht,hc,n); // 构造Huffman树，并形成字符的编码

for(i=1; i<=n; i++)

cout<<hc[i].data<<"---"<<hc[i].code<<endl;

print(ht, 2*n-1, 0);
enCoding(ht,hc,n);
cout<<endl; 
 return 0;  

}

Answer 1

http://wenku.baidu.com/link?url=cPLkMlJI-ZF2VGbEDHs3cVABOiZy_JrtRFdLYfCdPH7oOep4Pq9Lqpc2Gk-b6FoNpoxVZmBjr53rhGQ8QJu1qgyijpyM1CF3mn9KHLCji5O