哈夫曼树的c语言实现

哈夫曼树的c语言实现

#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 codes
void 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;
}