#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define N 50//叶子结点数
#define M 2*N-1//数中结点总数
typedef struct
{
char data[5];//结点值
int weight;//权重
int parent;//双亲结点
int lchild;//左孩子结点
int rchild;//右孩子结点
}HTNode;
typedef struct
{
char cd[N];//存放哈夫曼码
int start;
}HCode;
void CreateHT(HTNode ht[],int n)//构造哈夫曼树
{
int i,k,lnode,rnode;
int min1,min2;
for(i=0;i<2*n-1;i++)//所有结点的相关域置初值-1
ht[i].parent=ht[i].lchild=ht[i].rchild=-1;
for(i=n;i<2*n-1;i++)//构造哈夫曼树
{
min1=min2=32767;//lnode和rnode为最小权重的两个结点位置
lnode=rnode=-1;
for(k=0;k<=i-1;k++)
if(ht[k].parent==-1)
{
if(ht[k].weight<min1)//只在尚未构造二叉树的结点中查找
{
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++)//根据哈夫曼树求哈夫曼编码
{
hc.start=n;
c=i;
f=ht[i].parent;
while(f!=-1)//循序知道树根结点
{
if(ht[f].lchild==c)//处理左孩子结点
hc.cd[hc.start--]='0';
else//处理右孩子结点
hc.cd[hc.start--]='1';
c=f;
f=ht[f].parent;
}
hc.start++;//start指向哈夫曼编码最开始字符
hcd[i]=hc;
}
}
void DispHCode(HTNode ht[],HCode hcd[],int n)//输出哈夫曼编码
{
int i,k;
int sum=0,m=0,j;
printf(" 输出哈夫曼编码:\n");//输出哈夫曼编码
for(i=0;i<n;i++)
{
j=0;
printf("%s: ",ht[i].data);
for(k=hcd[i].start;k<=n;k++)
{
printf(" %c",hcd[i].cd[k]);
j++;
}
m+=ht[i].weight;
sum+=ht[i].weight*j;
printf("\n");
}
printf("\n 平均长度=%g\n",1.0*sum/m);
}
void main()
{
int n=15,i;
char*str[]={"The","of","a","to","and","in","that","he","is","at","on","for","His","are","be"};
int fnum[]={1192,677,541,518,462,450,242,195,190,181,174,157,138,124,123};
HTNode ht[M];
HCode hcd[N];
for(i=0;i<n;i++)
{
strcpy(ht[i].data,str[i]);
ht[i].weight=fnum[i];
}
printf("\n");
CreateHT(ht,n);
CreateHCode(ht,hcd,n);
DispHCode(ht,hcd,n);
printf("\n");
}