Huffman编码/译码问题
作者：互联网
Huffman编码/译码
问题描述
利用哈夫曼编码进行信息通讯可以大大提高信道利用率，缩短信息传输时间，降低传输成本。但是，这要求在发送端通过一个编码系统对待传数据预先编码；在接收端将传来的数据进行译码(复原)。对于双工信道(即可以双向传输信息的信道)，每端都需要一个完整的编/译码系统。试为这样的信息收发站写一个哈夫曼码的编译码系统。
基本要求
一个完整的系统应具有以下功能:
(1) I:初始化(Initialization)。从终端读入字符集大小n，及n个字符和n个权值，建立哈夫曼树，并将它存于文件hfmtree中。
(2) C:编码(Coding)。利用已建好的哈夫曼树(如不在内存，则从文件hfmtree中读 入)，对文件tobetrans中的正文进行编码，然后将结果存入codefile中。
(3) D:译码(Decoding)。利用已建好的哈夫曼树将文件codefile中的代码进行译 码，结果存入文件textfile中。
(4) P:印代码文件(Print)。将文件codefile以紧凑格式显示在终端上，每行50个代 码。同时将此字符形式的编码文件写入文件codeprint中。
(5) T:印哈夫曼树(Tree print)。将已在内存中的哈夫曼树以直观的方式 树或凹入表行式)显示在终端上，同时将此字符形式的哈夫曼树写入文件treeprint中。
测试数据：
用下表中给出的字符集和频度的实际统计数据建立哈夫曼树，并实现以下报文的编码和译码: “THIS PROGRAM IS MY FAVORITE”.
实现提示
(1) 文件codefile的基类型可以设为子界型 bit=0…1。
(2) 用户界面可以设计为"菜单"方式: 显示上述功能符号，再加上"E"，表示结束运行End，请用户键入一个选择功能符。此功能执行完毕后再显示此菜单，直至某次用户选择了"E"为止。
(3) 在程序的一次执行过程中，第一次执行I，D或C命令之后，哈符曼树已经在内存了，不必再读入。每次执行中不一定执行I命令，因为文件hfmtree可能早己建好。
#include <stdio.h>
#include <stdbool.h>
#include <stdlib.h>
#include <string.h>
typedef struct HT{
int val;
char word;
bool bitt[30];
struct HT *left, *right;
};
struct HT *Huffman_Tree[30], *roott;
bool Huff[30][30];
char Huffq[30][30];
int nn = 30, HuffLen[30];
void Build_Tree(int n, struct HT *root, int k) {
struct HT *q = (struct HT*) malloc(sizeof(struct HT));
q->left = NULL;
q->right = NULL;
q->val = 0;
if (Huffq[n][k] == '\0') {
if (n == 26) {
root->word = ' ';
}
else {
root->word = n + 'A';
}
}
else if (Huffq[n][k] == '0') {
if (root->left == NULL) {
q->word = '*';
root -> left = q;
Build_Tree(n, q, k + 1);
}
else {
Build_Tree(n, root->left, k + 1);
}
}
else if (Huffq[n][k] == '1') {
if (root->right == NULL) {
q->word = '*';
root -> right = q;
Build_Tree(n, q, k + 1);
}
else {
Build_Tree(n, root->right, k + 1);
}
}
}
void Readhfmtree() {
char s[30];
FILE *fp = NULL;
fp = fopen("../hfmtree.txt", "r");
if (fp == NULL) {
printf("The document does not exist\n");
}
int sum = 0;
while(!feof(fp)) {
memset(s, '\0', sizeof(s));
fgets(s, 30, fp);
for (int i = 2; i < strlen(s); i++) {
if (s[i] == '\n') {
break;
}
Huffq[sum][i - 2] = s[i];
}
sum++;
}
fclose(fp);
roott = (struct HT*) malloc(sizeof(struct HT));
roott->left = NULL;
roott->right = NULL;
roott->word = '*';
roott->val = 0;
for (int i = 0; i < 27; i++) {
Build_Tree(i, roott, 0);
}
}
void sortt(int n) {
for (int i = 0; i < n - 1; i++) {
for(int j = i + 1; j < n; j++) {
if (Huffman_Tree[i]->val < Huffman_Tree[j]->val) {
struct HT *temp = Huffman_Tree[i];
Huffman_Tree[i] = Huffman_Tree[j];
Huffman_Tree[j] = temp;
}
}
}
}
void Build_Bitt(struct HT *root, int k) {
if (root == NULL) {
return;
}
if (root->word >= 'A' && root->word <= 'Z') {
memcpy(Huff[root->word - 'A'], root->bitt, k + 1);
HuffLen[root->word - 'A'] = k;
}
else if (root->word == ' ') {
memcpy(Huff[26], root->bitt, k + 1);
HuffLen[26] = k;
}
if (root -> left != NULL) {
memcpy(root->left->bitt, root->bitt, k + 1);
root->left->bitt[k + 1] = false;
Build_Bitt(root->left, k + 1);
}
if (root -> right != NULL) {
memcpy(root->right->bitt, root->bitt, k + 1);
root->right->bitt[k + 1] = true;
Build_Bitt(root->right, k + 1);
}
}
void Initialization() {
char ch;
int q;
getchar();
printf("input the letter and its val\n");
for (int i = 0; i < 27; ++i) {
scanf("%c %d", &ch, &q);
struct HT *ht = (struct HT*)malloc(sizeof(struct HT));
ht->word = ch;
ht->val = q;
ht->left = NULL;
ht->right = NULL;
memset(ht->bitt, false, sizeof(ht->bitt));
getchar();
if (ch == ' ') {
Huffman_Tree[26] = ht;
}
else {
Huffman_Tree[ch - 'A'] = ht;
}
}
sortt(27);
int m = 27;
while(m != 1) {
struct HT *ht = (struct HT *)malloc(sizeof(struct HT));
ht->left = Huffman_Tree[m - 2];
ht->right = Huffman_Tree[m - 1];
ht->word = '*';
ht->val = ht->left->val + ht->right->val;
memset(ht->bitt, false, sizeof(ht->bitt));
Huffman_Tree[m - 2] = ht;
sortt(m);
m--;
}
Build_Bitt(Huffman_Tree[0], 0);
FILE *fp = NULL;
fp = fopen("../hfmtree.txt","w");
if (fp == NULL) {
printf("Document Error");
}
for (int i = 0; i < 27; i++) {
char s[30];
memset(s, '\0', sizeof(s));
if (i == 27 - 1) {
s[0] = ' ';
}
else {
s[0] = 'A' + i;
}
s[1] = ' ';
for (int j = 1; j <= HuffLen[i]; ++j) {
if (Huff[i][j] == false) {
s[j + 1] = '0';
}
else {
s[j + 1] = '1';
}
}
fprintf(fp, s);
fprintf(fp, "\n");
}
fclose(fp);
}
void Coding() {
char s[300];
memset(s, '\0', sizeof(s));
gets(s);
FILE *fp = NULL;
fp = fopen("../codefile.txt","w");
if (fp == NULL) {
printf("Document Error");
}
for (int i = 0; i < strlen(s); i++) {
if (s[i] >= 'A' && s[i] <= 'Z') {
fprintf(fp, Huffq[s[i] - 'A']);
}
else if (s[i] == ' ') {
fprintf(fp, Huffq[26]);
}
else {
char ch[2];
ch[0] = s[i];
ch[1] = '\0';
fprintf(fp, ch);
}
}
fprintf(fp, "\n");
fclose(fp);
};
void Decoding() {
char s1[3000], s2[3000];
FILE *fp1 = NULL, *fp2 = NULL;
fp1 = fopen("../codefile.txt","r");
fp2 = fopen("../textfile.txt","w");
if (fp2 == NULL) {
printf("Document Error");
}
if (fp1 == NULL) {
printf("Document Not exist");
}
while (!feof(fp1)) {
memset(s1, '\0', sizeof(s1));
memset(s2, '\0', sizeof(s2));
fgets(s1, 3000, fp1);
struct HT *q = roott;
int sum = 0;
for (int i = 0; i < strlen(s1); ++i) {
if (s1[i] == '0') {
q = q->left;
}
else if (s1[i] == '1') {
q = q->right;
}
else {
s2[sum++] = s1[i];
}
if (q->word != '*') {
s2[sum++] = q->word;
q = roott;
}
}
fprintf(fp2, s2);
fprintf(fp2, "\n");
}
fclose(fp2);
fclose(fp1);
}
void Print() {
char s[1000];
FILE *fp = NULL;
fp = fopen("../codefile.txt","r");
if (fp == NULL) {
printf("Document Not exist");
}
while(!feof(fp)) {
memset(s, '\0', sizeof(s));
fgets(s, 1000, fp);
for (int i = 0; i < strlen(s); ++i) {
printf("%c", s[i]);
if ((i + 1) % 50 == 0) {
printf("\n");
}
}
}
fclose(fp);
}
void Tree_Print() {
FILE *fp;
fp = fopen("../treeprint.txt","w");
if (fp == NULL) {
printf("Document Error");
}
int sum = 30;
char s[50];
for (int i = 0; i < 27; ++i) {
memset(s, '\0', sizeof(s));
if (i == 26) {
s[0] = ' ';
}
else {
s[0] = 'A' + i;
}
s[1] = ' ';
for (int j = 0; j < sum - strlen(Huffq[i]) * 2; ++j) {
s[j + 2] = '=';
}
fprintf(fp, s);
fprintf(fp,"\n");
printf("%s\n",s);
}
fclose(fp);
}
int main() {
char c;
while (1) {
printf(" 1.Initialization\n 2.Coding\n 3.Decoding\n 4.Print\n 5.Tree print\n");
printf("input the number of the instruction to use the instructions, or input e/E to exit\n");
scanf("%c",&c);
if (c == 'E' || c == 'e') {
break;
}
else {
if (c == '1') {
Initialization();
}
else if (c == '2') {
Readhfmtree();
getchar();
Coding();
}
else if (c == '3') {
Readhfmtree();
Decoding();
}
else if (c == '4') {
Print();
}
else if (c == '5') {
Readhfmtree();
Tree_Print();
}
}
}
return 0;
}
标签：fp,编码,int,root,Tree,else,译码,NULL,Huffman 来源： https://blog.csdn.net/transbit/article/details/122549833