📄 哈夫曼树.cpp
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#include<iostream.h>
#include<string.h>
#include<stdio.h>
#define N 10
#define M 2*N-1
typedef struct
{
int weight;
int parent,lchild,rchild;
}htnode; //树中结点的结构
typedef struct
{
char data;
int weight; //字符的权值
char code[N];
}htcode;
void init(htcode hc[],int n)
{//初始化,读入编码字符的个数,从键盘输入6个字符和6个权值
int i;
cout<<"输入 "<<n<<" 个叶子结点: "<<endl;
for(i=1;i<=n;i++)
cin>>hc[i].data;
cout<<"依次输入 "<<n<<" 个叶子结点的权值:"<<endl;
for(i=1;i<=n;i++)
cin>>hc[i].weight;
}
void select(htnode ht[],int n,int & s1,int & s2)
{//ht[1....k]中选择parent为0,并且weight最小的两个结点,
//其序号由变量s1,s2返回
int i;
int max;
int flag=1;
for(i=1;i<=n;i++)
{ //找前两个父结点为0的结点,用s1,s2返回其序号
if(ht[i].parent==0 )
{
if(flag)
{s1=i;flag=0;}
else
{s2=i;break;}
}
}
for(i=s2+1;i<=n;i++)
{
max=ht[s1].weight>ht[s2].weight?ht[s1].weight:ht[s2].weight; //max=权大的结点的权
if(ht[i].parent==0 && ht[i].weight<max)
if(ht[s1].weight>ht[s2].weight)s1=i; //因为ht[s1].weight>ht[s2].weight,
//并且ht[i].weight<ht[s1].weight,所以使s1=i
else s2=i;
} //// 寻找权值最小的两结点
}
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];
int i,m,c,f,s1,s2,start;
m=2*n-1; //n=待编码字符个数 有N个叶子结点的二叉树有2n-1个结点
for(i=1;i<=m;i++) //初始化huffmantree
{
if(i<=n)
ht[i].weight=hc[i].weight;
else
ht[i].weight=0;
ht[i].parent=ht[i].lchild=ht[i].rchild=0;
}
for(i=n+1;i<=m;i++)////建立huffmantree
{
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!=0;c=f,f=ht[f].parent)
if(ht[f].lchild==c)
cd[--start]='0';
else
cd[--start]='1';
strcpy(hc[i].code,&cd[start]);////huffman编码
}
}
void main()
{
int i,n=6;
htnode ht[20];
htcode hc[11];
init(hc,n);
huffmancoding(ht,hc,n);
cout<<"huffman编码为:"<<endl;
for(i=1;i<=n;i++)
cout<<hc[i].data<<"---------"<<hc[i].code<<endl;
}
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