数据结构——构成哈弗曼树的问题

<span style="font-family:SimHei;font-size:24px;color:#ff0000;"><em>#include <stdio.h>
#include <iostream>
#include <string>
using namespace std;

#define N 50        //叶子结点数
#define M 2*N-1     //树中结点总数</em></span>

//哈夫曼树的节点结构类型


<span style="font-size:18px;"><strong>typedef struct
{
    string data;  //结点值
    double weight;  //权重
    int parent;     //双亲结点
    int lchild;     //左孩子结点
    int rchild;     //右孩子结点
} HTNode;
</strong></span>

//每个节点哈夫曼编码的结构类型

<strong><span style="font-size:18px;">typedef struct
{
    char cd[N]; //存放哈夫曼码
    int start;
} HCode;</span></strong>

//构造哈夫曼树

void CreateHT(HTNode ht[],int n)
{
    int i,k,lnode,rnode;
    double 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)   //只在尚未构造二叉树的结点中查找
            {
                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;
    }
}

//实现哈夫曼编码

void CreateHCode(HTNode ht[],HCode hcd[],int n)
{
    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)   //循序直到树根结点
        {
            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;
    }
}

//输出哈夫曼编码

void DispHCode(HTNode ht[],HCode hcd[],int n)
{
    int i,k;
    double sum=0,m=0;
    int 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("      %c:\t",ht[i].data);  //vc++ 6.0中会出错,主要原因是因为该平台中c与c++不兼容。。
		cout<<ht[i].data<<'\t';//同时运用printf与out时同时写上#include <stdio.h>  #include <iostream>就可以了
        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);
}


主函数

int main()
{
    int n=15,i;      //n表示初始字符串的个数
    string str[15]= {"The:","of:","a:","to:","and:","in:","that:","he:","is","at:","on:","for:","His:","are:","be:"};
    double 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++)
    {
        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");
    return 0;
}


效果图:

                                          《数据结构——构成哈弗曼树的问题》



    原文作者:哈夫曼树
    原文地址: https://blog.csdn.net/alim1314520n/article/details/51683671
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞