数据结构第七次试验

一、实验题目
设计一个程序exp7-6.cpp，构造一棵哈夫曼树，输出对应的哈夫曼编码和平均查找长度。并用表7.8所示的数据进行验证。
表7.8 单词及出现的频度

单词	The	of	a	to	and	in	that	he	is	at	on	for	His	are	be
出现频度	1192	677	541	518	462	450	242	195	190	181	174	157	138	124	123

 
二、实验目的
掌握哈夫曼树的构造过程和哈夫曼编码的产生方法；灵活运用二叉树这种数据结构解决一些综合应用问题。
三、实验要求
针对程序exp7-6.cpp，输出结果如下：
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
#define N 50
#define M 2*N-1
using namespace std;
typedef struct
{
string data;
double weight;
int parent;
int lchild;
int rchild;
}HTNode;
typedef struct
{
char cd
;
int start;
}HCode;
void CreateHT(HTNode ht[],int n)
{
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=32767;
lnode=rnode=-1;
for(k=0;k<=i-1;k++)
{
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[i].weight=ht[lnode].weight+ht[rnode].weight;
ht[i].lchild=lnode;ht[i].rchild=rnode;
ht[lnode].parent=i;ht[rnode].parent=i;
}
}
void DispHCode(HTNode ht[],HCode hcd[],int n)
{
int i,k;
int sum=0,m=0,j;
cout<<"输出哈夫曼编码:"<<endl; //输出哈夫曼编码
for (i=0;i<n;i++)
{
j=0;
cout<<ht[i].data<<'\t';
for (k=hcd[i].start;k<=n;k++)
{
cout<<hcd[i].cd[k];
j++;
}
m+=ht[i].weight;
sum+=ht[i].weight*j;
cout<<endl;
}
cout<<"平均长度=";
cout<<1.0*sum/m;
}
void CreateHCode(HTNode ht[],HCode hcd[],int n)
{
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)
hc.cd[hc.start--]='0';
else
hc.cd[hc.start--]='1';
c=f;f=ht[f].parent;
}
hc.start++;
hcd[i]=hc;
}
}

int main()
{
int n=15,i;
string 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
;
for (i=0;i<n;i++)
{
ht[i].data=str[i];
ht[i].weight=fnum[i];
}
CreateHT(ht,n);
CreateHCode(ht,hcd,n);
DispHCode(ht,hcd,n);
return 0;
}