哈夫曼树的编码实现

2024-01-20 10:02:28

#include<stdio.h>

#include<string.h>
#include<stdlib.h>
//树结点定义
typedef struct
{
int weight;
int parent;
int lchild;
int rchild;
}HTNode,*HuffmanTree;
static char N[100];//用于保存正文
//哈弗曼编码，char型二级指针
typedef char **HuffmanCode;
//封装最小权结点和次小权结点
typedef struct
{
int s1;
int s2;
}MinCode;
//函数声明
void Error(char *message);
HuffmanCode HuffmanCoding(HuffmanTree &HT,HuffmanCode HC,int *w,int n);
MinCode Select(HuffmanTree HT,int n);
//当输入1个结点时的错误提示
void Error(char *message)
{
fprintf(stderr,"Error:%s\n",message);
exit(1);
}
//构造哈夫曼树HT，编码存放在HC中,w为权值,n为结点个数
HuffmanCode HuffmanCoding(HuffmanTree &HT,HuffmanCode HC,int *w,int n)
{
int i,s1=0,s2=0;
HuffmanTree p;
char *cd;
int f,c,start,m;
MinCode min;
if(n<=1)
{
Error("Code too small!");//只有一个结点不进行编码，直接exit(1)退出。非return,如果return 会造成main函数HT[i]无值
}
m=2*n-1;//哈弗曼编码需要开辟的结点大小为2n-1
HT=(HuffmanTree)malloc((m+1)*sizeof(HTNode));//开辟哈夫曼树结点空间 m+1 。为了对应关系，我们第0个空间不用。
//初始化n个叶子结点,w[0] = 0,main函数已赋值
for(p=HT,i=0;i<=n;i++,p++,w++)
{
p->weight=*w;
p->parent=0;
p->lchild=0;
p->rchild=0;
}
//将n-1个非叶子结点的初始化
for(;i<=m;i++,p++)
{
p->weight=0;
p->parent=0;
p->lchild=0;
p->rchild=0;
}
//构造哈夫曼树
for(i=n+1;i<=m;i++)
{
min=Select(HT,i-1);//找出最小和次小的两个结点
s1=min.s1 ; //最小结点下标
s2=min.s2;//次小结点下标
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;//赋权和
}
//打印哈弗曼树
printf("HT List:\n");
printf("Number\t\tweight\t\tparent\t\tlchild\t\trchild\n");
for(i=1;i<=m;i++)
{
printf("%d\t\t%d\t\t%d\t\t%d\t\t%d\t\n",i,HT[i].weight,HT[i].parent,HT[i].lchild,HT[i].rchild);
}
//从叶子结点到根节点求每个字符的哈弗曼编码
HC=(HuffmanCode)malloc((n+1)*sizeof(char *));
cd=(char *)malloc(n*sizeof(char *));//为哈弗曼编码动态分配空间
cd[n-1]='\0';//如：3个结点编码最长为2。cd[3-1] = '\0';
//求叶子结点的哈弗曼编码
for(i=1;i<=n;i++)
{
start=n-1;
//定义左子树为0，右子树为1
/*
从最下面的1号节点开始往顶部编码(逆序存放)，然后编码2号节点，3号......
*/
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';
}
//为第i个字符分配编码空间
HC[i]=(char *)malloc((n-start)*sizeof(char *));
//将当前求出结点的哈弗曼编码复制到HC
strcpy(HC[i],&cd[start]);
}
free(cd);
return HC;
}
MinCode Select(HuffmanTree HT,int n)
{
int min,secmin;
int temp = 0;
int i,s1,s2,tempi = 0;
MinCode code ;
s1=1;
s2=1;
min = 66666;//足够大
//找出权值weight最小的结点，下标保存在s1中
for(i=1;i<=n;i++)
{
if(HT[i].weight<min && HT[i].parent==0)
{
min=HT[i].weight;
s1=i;
}
}
secmin = 66666;//足够大
//找出权值weight次小的结点，下标保存在s2中
for(i=1;i<=n;i++)
{
if((HT[i].weight<secmin) && (i!=s1) && HT[i].parent==0)
{
secmin=HT[i].weight;
s2=i;
}
}
//放进封装中
code.s1=s1;
code.s2=s2;
return code;
}
void HuffmanTranslateCoding(HuffmanTree HT, int n,char* ch)
{//译码过程
int m=2*n-1;
int i,j=0;
printf("After Translation:");
while(ch[j]!='\0')//ch[]:你输入的要译码的0101010串
{
i=m;
while(0 != HT[i].lchild && 0 != HT[i].rchild)//从顶部找到最下面
{
if('0' == ch[j])//0 往左子树走
{
i=HT[i].lchild;
}
else//1 往右子树走
{
i=HT[i].rchild;
}
++j;//下一个路径
}
printf("%c",N[i-1]);//打印出来
}
printf("\n");
}
void main()
{
HuffmanTree HT=NULL;
HuffmanCode HC=NULL;
int *w=NULL;
int i,n;
char tran[100];
printf("Input N(char):");
gets(N);
fflush(stdin);
n = strlen(N);
w=(int *)malloc((n+1)*sizeof(int *));//开辟n+1个长度的int指针空间
w[0]=0;
printf("Enter weight:\n");
//输入结点权值
for(i=1;i<=n;i++)
{
printf("w[%d]=",i);
scanf("%d",&w[i]);
}
fflush(stdin);
//构造哈夫曼树HT，编码存放在HC中,w为权值,n为结点个数
HC=HuffmanCoding(HT,HC,w,n);
//输出哈弗曼编码
printf("HuffmanCode:\n");
printf("Number\t\tWeight\t\tCode\n");
for(i=1;i<=n;i++)
{
printf("%c\t\t%d\t\t%s\n",N[i-1],w[i],HC[i]);
}
fflush(stdin);
//译码过程
printf("Input HuffmanTranslateCoding:");
gets(tran);
HuffmanTranslateCoding(HT, n, tran);
return;
}

#include <stdio.h>#include <stdlib.h>#include <memory.h> #define MAX_CODE_LENGTH 40 long long vocab_max_size = 1000, vocab_size = 0, layer1_size = 100; /** * word与Huffman树编码 */struct vocab_word { long long cn; // 词在训练集中出现的频率 int *point; // 编码的节点路径 char *word, // 词 *code, // Huffman编码，每一位上，0或1 codelen; // Huffman编码长度}; struct vocab_word *vocab; /* * 打印构造过程的中间状态. */ void printState(long long* count, long long* binary, long long* parent_node) { printf("count[]:\t"); for(int x=0; x<vocab_size * 2; x++) { printf("%lld", count[x]); printf(" "); } printf("\n"); printf("binary[]:\t"); for(int x=0; x<vocab_size * 2; x++) { printf("%lld", binary[x]); printf(" "); } printf("\n"); printf("parent[]:\t"); for(int x=0; x<vocab_size * 2; x++) { printf("%lld", parent_node[x]); printf(" "); } printf("\n");}/** * 使用词频创建一棵的Huffman树. 频率高的字将具有更短的 * Huffman二进制码(binary code). * */// Create binary Huffman tree using the word counts// Frequent words will have short uniqe binary codesvoid CreateBinaryTree() { long long a, b, i, min1i, min2i, pos1, pos2, point[MAX_CODE_LENGTH]; char code[MAX_CODE_LENGTH]; // count: 词频. // binary: // parent_node: long long *count = (long long *)calloc(vocab_size * 2 + 1, sizeof(long long)); long long *binary = (long long *)calloc(vocab_size * 2 + 1, sizeof(long long)); long long *parent_node = (long long *)calloc(vocab_size * 2 + 1, sizeof(long long)); // 初始化count数组的前一半 for (a = 0; a < vocab_size; a++) { count[a] = vocab[a].cn; //printf("count=%lld\n", count[a]); } // 初始化count数组的后一半，用于交换. 赋很大值. for (a = vocab_size; a < vocab_size * 2; a++) { count[a] = 1e15; //printf("count=%lld\n", count[a]); } // pos1 = vocab_size - 1; pos2 = vocab_size; //printf("pos1=%lld, pos2=%lld\n", pos1, pos2); printState(count, binary, parent_node); // 根据算法构建Huffman树，一次增加一个节点. // Following algorithm constructs the Huffman tree by adding one node at a time for (a = 0; a < vocab_size - 1; a++) { printf("----------------\n"); printf("pos1=%lld, pos2=%lld\n", pos1, pos2); // 每轮找到最小的两个值. // First, find two smallest nodes 'min1, min2' if (pos1 >= 0) { // 遍历所有词汇的count，比较count；取较小值. if (count[pos1] < count[pos2]) { min1i = pos1; pos1--; } else { min1i = pos2; pos2++; } } else { min1i = pos2; pos2++; } printf("min1i=%d, min2i=%d\n", min1i, min2i); printf("pos1=%lld, pos2=%lld\n", pos1, pos2); // 再比一次. if (pos1 >= 0) { if (count[pos1] < count[pos2]) { min2i = pos1; pos1--; } else { min2i = pos2; pos2++; } } else { min2i = pos2; pos2++; } // 最小值cnt的两个索引 printf("min1i=%d, min2i=%d\n", min1i, min2i); printf("count[min1i]=%d, count[min2i]=%d\n", count[min1i], count[min2i]); count[vocab_size + a] = count[min1i] + count[min2i]; parent_node[min1i] = vocab_size + a; parent_node[min2i] = vocab_size + a; binary[min2i] = 1; printf("count[vocab_size + a] = %d\n", count[vocab_size + a]); printf("parent_node[%d] = %d\n", min1i, parent_node[min1i]); printf("parent_node[%d] = %d\n", min2i, parent_node[min2i]); printf("binary[%d] = %d\n", min2i, binary[min2i]); printState(count, binary, parent_node); } // 将二进制编码分配给词汇表中每个词汇. // Now assign binary code to each vocabulary word for (a = 0; a < vocab_size; a++) { b = a; i = 0; while (1) { code[i] = binary[b]; point[i] = b; i++; b = parent_node[b]; if (b == vocab_size * 2 - 2) break; } // 得到huffman编码长度. vocab[a].codelen = i; // 得到huffman编码code及路径point. vocab[a].point[0] = vocab_size - 2; for (b = 0; b < i; b++) { vocab[a].code[i - b - 1] = code[b]; vocab[a].point[i - b] = point[b] - vocab_size; } } // 释放内存. free(count); free(binary); free(parent_node);}/** * 代码运行： * gcc ./huffman_tree.cpp; ./a.out */int main(){ vocab_size = 6; vocab = (vocab_word*) calloc(vocab_size, sizeof(vocab_word)); memset(vocab, 0, sizeof(vocab_word) * (vocab_size)); // 初始化code/point. for (int a = 0; a < vocab_size; a++) { vocab[a].code = (char *)calloc(MAX_CODE_LENGTH, sizeof(char)); vocab[a].point = (int *)calloc(MAX_CODE_LENGTH, sizeof(int)); } // 事先对vocab按词频排好序(word2vec事先已经用qsort处理)，从大到小排序. // 可以用qsort。本代码直接已经人工排好序了. vocab[0].cn = 7; char* str = "T"; vocab[0].word = str; vocab[1].cn = 5; str = "E"; vocab[1].word = str; vocab[2].cn = 4; str = "G"; vocab[2].word = str; vocab[3].cn = 4; str = "R"; vocab[3].word = str; vocab[4].cn = 3; str = "O"; vocab[4].word = str; vocab[5].cn = 2; str = "F"; vocab[5].word = str; CreateBinaryTree(); for (int a = 0; a < vocab_size; a++) { printf("word=%s\t", vocab[a].word); printf("cn=%d\t", vocab[a].cn); printf("codelen=%d\t", vocab[a].codelen); printf("code="); for(int i = 0; i < vocab[a].codelen; i++) { printf("%d", vocab[a].code[i]); } printf("\t"); printf("point="); for(int i = 0; i < vocab[a].codelen; i++) { printf("%d-", vocab[a].point[i]); } printf("\n"); //printf("point=%s\n", vocab[a].point); } return 0;
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版权声明：本文为CSDN博主「桃根仙」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/taotaobaobei/java/article/details/78513979

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