Solve for Huffman encoding.

This is because everything in your program involves tree[i].weight, where both input and output are %d, and the data type you define weight is float, so there is a type mismatch, and the data you input is not stored in tree[i].weight variable, as for your own algorithm, I didn't look closely.

The program is modified as follows:

#include

#define maxsize 30

typedef struct

float weight;

char flag;

int lchild;

int rchild;

int parent;

tree;main()

int n,i,m,p1,p2,j,k,number;

float minvalue1,minvalue2;

tree tree[maxsize];

printf("Please enter the number of nodes");

scanf("%d",&n);

m=2*n;

for(i=0;itree[j].weight))minvalue1=tree[j].weight;

p1=j;tree[p1].flag=1;

for(k=1;tree[k].flag!=0;k++)minvalue2=tree[k].weight;

p2=k;for(j=k+1;j<=number;j++)tree[j].weight))

minvalue2=tree[j].weight;

p2=j;tree[p1].parent=i;

tree[p1].flag=1;

tree[p2].parent=i;

tree[p2].flag=1;

tree[i].lchild=p1;

tree[i].rchild=p2;

tree[i].weight=tree[p1].weight+tree[p2].weight;

for(i=1;i<2*number;i++)printf("%5f",tree[i].weight);Pay attention here!

printf("");

for(i=1;Attention here!

I feel dizzy when I see such a long **!

#include

#include

using namespace std;

typedef struct

hnodetype;

typedef struct

hcodetype;

void huf(char cha,int m,int n)for(i=0;i>str;

n=;cout<<"Strings have a total of characters"continue;

for(j=i;jcoutk++;cout<<"The Huffman encoding for each character is:"

Changed, is that okay?

Enter the number of characters:

4 Enter each character and weight:

a 4b 3

c 2d 4

a: 4: 10

b: 3: 01

c: 2: 00

d: 4: 11

4 84 Enter the code: (Note that please enter the standard 01 code here, no other inspection) 10010011

ABCD press any key to continue. .

First of all, only when constructing the tree may appear the same value, but that doesn't matter, which can be taken out by the way, but even after the formation of the same code will definitely not appear, hafuman is to solve the same value coding problem, **a tree from the root to any**leaf**The path is unique,

First, analyze the weights of each character:

a=3，b＝7，c＝2，d＝3，e＝5

Generate a Hoffman tree and get the encoding of each character:

A = 110, b = 0, c = 1111, d = 1110, e = 10 The average code length is 46 15

Hoffman coding has some distinct features:

1) The coded codes are all different prefix codes, which ensures the unique translatability of the codes.

2) Due to the variable length of the code. As a result, the decoding time is long, which makes the compression and restoration of Hoffman encoding quite time-consuming.

3) The coding length is not uniform, and the hardware implementation is difficult.

4) The coding efficiency of different signal sources is different, when the sign probability of the signal source is negative power of 2, the coding efficiency reaches 100%; If the probabilities of the signal source symbols are equal, the encoding efficiency is the lowest.

5) Due to"0"with"1"Therefore, the optimal code coded by the above process is not unique, but its average code length is the same, so it does not affect the encoding efficiency and data compression performance.

Huffman encoding is a frequency-dependent encoding, first you have to count the frequency of the text in your rolling text delay file, because the content of the text file contains Chinese, so when read with c can only be counted according to bytes, because the bytes in the machine only have a value range of 0 to 255, you can determine that the nodes of the huffman tree we want to build are 256, and the statistical frequency can be solved by using the map container of C++.