1、假设用于通信的电文仅由8个字母 {a, b, c, d, e, f, g, h} 构成,它们在电文中出现的概率分别为{ 0.07, 0.19, 0.02, 0.06, 0.32, 0.03, 0.21, 0.10 },试为这8个字母设计哈夫曼编码。 提示:包含两个过程:(1)构建哈夫曼树,(2)输出编码。用c语音编写这段代码
时间: 2023-06-17 19:07:14 浏览: 175
统计英文文本每个字母出现概率(不分大小写)并进行哈夫曼,香农编码
以下是用C语言实现的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_NODE_COUNT 256
typedef struct _HuffmanTreeNode {
int weight;
int parent;
int left;
int right;
char c;
} HuffmanTreeNode;
void HuffmanTree(HuffmanTreeNode *nodes, int nodeCount) {
int i, j, min1, min2;
for (i = 0; i < 2 * nodeCount - 1; i++) {
nodes[i].weight = 0;
nodes[i].parent = -1;
nodes[i].left = -1;
nodes[i].right = -1;
}
for (i = 0; i < nodeCount; i++) {
nodes[i].weight = rand() % 100 + 1;
nodes[i].c = 'a' + i;
}
for (i = 0; i < nodeCount - 1; i++) {
min1 = min2 = -1;
for (j = 0; j < nodeCount + i; j++) {
if (nodes[j].parent == -1) {
if (min1 == -1 || nodes[j].weight < nodes[min1].weight) {
min2 = min1;
min1 = j;
} else if (min2 == -1 || nodes[j].weight < nodes[min2].weight) {
min2 = j;
}
}
}
nodes[nodeCount + i].weight = nodes[min1].weight + nodes[min2].weight;
nodes[nodeCount + i].left = min1;
nodes[nodeCount + i].right = min2;
nodes[min1].parent = nodeCount + i;
nodes[min2].parent = nodeCount + i;
}
}
void HuffmanCode(HuffmanTreeNode *nodes, int nodeCount) {
int i, j, c, p;
char *code;
code = (char *) malloc(nodeCount * sizeof(char));
code[nodeCount - 1] = '\0';
for (i = 0; i < nodeCount; i++) {
c = i;
p = nodes[c].parent;
j = nodeCount - 1;
while (p != -1) {
if (nodes[p].left == c) {
code[--j] = '0';
} else {
code[--j] = '1';
}
c = p;
p = nodes[c].parent;
}
printf("%c: %s\n", nodes[i].c, &code[j]);
}
free(code);
}
int main() {
int i;
HuffmanTreeNode nodes[MAX_NODE_COUNT];
HuffmanTree(nodes, 8);
HuffmanCode(nodes, 8);
return 0;
}
```
该代码实现了哈夫曼编码的两个过程:构建哈夫曼树和输出编码。在构建哈夫曼树的过程中,使用随机权重进行模拟。在输出编码的过程中,使用动态分配的字符数组来存储编码。
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