怎么用C语言编程实现二元huffman,shannon,fano,S-F-E编码

追问：请问还有shannon,fano,S-F-E编码吗？谢谢

答案：huffman:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define LEN sizeof(struct HTnode)
int i,l,n,w=0,c,start,a1,a2,f;
struct HTnode {
 unsigned int weight;
 unsigned int parent,lchild,rchild;
       }*p,*HT;
typedef char **Huffmancode;
Huffmancode HC;
char *cd;
select()
{
 int k=1,j,flag=0;
 while((HT+k)->parent!=0) k++;
 for(j=k+1;j<=n;j++,flag=0)
  {
   if((HT+j)->parent!=0) flag=1;
   if((HT+j)->weight==0) flag=1;
   if(!flag)
   {
    if((HT+j)->weight<(HT+k)->weight) k=j;
   }
  }
 return(k);
}
main()
{
 printf("\n赫夫曼树的建立:\n");
 printf("请输入权值(叶子)数目:");
 scanf("%d",&l);
 while(l<1) 
 {
  printf("输入错误,请重新输入权值数目:"); scanf("%d",&l);
 }
 if(l==1) printf("\n只有一个权值,无须建立赫夫曼树!");
 else 
 {
  n=2*l-1;
       HT=(struct HTnode*)malloc((n+1)*LEN);
       printf("请按对应顺序输入权值(输入一权值,键入一回车):\n");
       for(i=1,p=HT+1;i<=l;++i,++p)
 {
     scanf("%d",&w);
  while(w<=0){printf("权值错,重新输入此权值:"); scanf("%d",&w);
  }
  p->weight=w; p->parent=0;
  p->lchild=0; p->rchild=0;
 }
      for(i=l+1;i<=n;++i,++p)
       {
    p->weight=0; p->parent=0;
       p->lchild=0;
       }
      for(i=l+1;i<=n;++i)
       {
    a1=select(); (HT+a1)->parent=i;
       a2=select(); (HT+a2)->parent=i;
       (HT+i)->lchild=a1;
       (HT+i)->rchild=a2;
       (HT+i)->weight=(HT+a1)->weight+(HT+a2)->weight;
       }
     HC=(Huffmancode)malloc((l+1)*sizeof(char *));
     cd=(char *)malloc(l*sizeof(char));
     *(cd+(l-1))='\0';
     for(i=1;i<=l;++i)
      {
   start=l-1;
       for(c=i,f=(HT+i)->parent;f!=0;c=f,f=(HT+f)->parent)
       if((HT+f)->lchild==c) *(cd+(--start))='0';
       else *(cd+(--start))='1';
      *(HC+i)=(char *)malloc((l-start)*sizeof(char));
      strcpy(*(HC+i),(cd+start));
     }
    printf("\n对应的二进制赫夫曼编码为:\n");
    for(i=1;i<=l;++i)
      {
  printf("%s",*(HC+i));
       printf("\n");
      }
 }
}

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