用C语言编程实现赫夫曼编码 假定用于通信的电文由8个字母A、B、C、D、E、F、G、H组成，各字母在电文中出现的概率为5%，25%，4%，7%，9%，12%，30%，8%，试编程为这8个字母设计赫夫曼编码。

以下是用C语言实现赫夫曼编码的代码，其中包括了对于题目中给出的8个字母的概率进行赫夫曼编码的过程：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_TREE_HT 100

struct MinHeapNode {
    char data;
    unsigned freq;
    struct MinHeapNode *left, *right;
};

struct MinHeap {
    unsigned size;
    unsigned capacity;
    struct MinHeapNode **array;
};

struct MinHeapNode *newNode(char data, unsigned freq) {
    struct MinHeapNode *temp = (struct MinHeapNode *)malloc(sizeof(struct MinHeapNode));
    temp->left = temp->right = NULL;
    temp->data = data;
    temp->freq = freq;
    return temp;
}

struct MinHeap *createMinHeap(unsigned capacity) {
    struct MinHeap *minHeap = (struct MinHeap *)malloc(sizeof(struct MinHeap));
    minHeap->size = 0;
    minHeap->capacity = capacity;
    minHeap->array = (struct MinHeapNode **)malloc(minHeap->capacity * sizeof(struct MinHeapNode *));
    return minHeap;
}

void swapMinHeapNode(struct MinHeapNode **a, struct MinHeapNode **b) {
    struct MinHeapNode *t = *a;
    *a = *b;
    *b = t;
}

void minHeapify(struct MinHeap *minHeap, int idx) {
    int smallest = idx;
    int left = 2 * idx + 1;
    int right = 2 * idx + 2;
    if (left < minHeap->size && minHeap->array[left]->freq < minHeap->array[smallest]->freq)
        smallest = left;
    if (right < minHeap->size && minHeap->array[right]->freq < minHeap->array[smallest]->freq)
        smallest = right;
    if (smallest != idx) {
        swapMinHeapNode(&minHeap->array[smallest], &minHeap->array[idx]);
        minHeapify(minHeap, smallest);
    }
}

int isSizeOne(struct MinHeap *minHeap) {
    return (minHeap->size == 1);
}

struct MinHeapNode *extractMin(struct MinHeap *minHeap) {
    struct MinHeapNode *temp = minHeap->array[0];
    minHeap->array[0] = minHeap->array[minHeap->size - 1];
    --minHeap->size;
    minHeapify(minHeap, 0);
    return temp;
}

void insertMinHeap(struct MinHeap *minHeap, struct MinHeapNode *minHeapNode) {
    ++minHeap->size;
    int i = minHeap->size - 1;
    while (i && minHeapNode->freq < minHeap->array[(i - 1) / 2]->freq) {
        minHeap->array[i] = minHeap->array[(i - 1) / 2];
        i = (i - 1) / 2;
    }
    minHeap->array[i] = minHeapNode;
}

void buildMinHeap(struct MinHeap *minHeap) {
    int n = minHeap->size - 1;
    int i;
    for (i = (n - 1) / 2; i >= 0; --i)
        minHeapify(minHeap, i);
}

void printArr(int arr[], int n) {
    int i;
    for (i = 0; i < n; ++i)
        printf("%d", arr[i]);
    printf("\n");
}

int isLeaf(struct MinHeapNode *root) {
    return !(root->left) && !(root->right);
}

struct MinHeap *createAndBuildMinHeap(char data[], int freq[], int size) {
    struct MinHeap *minHeap = createMinHeap(size);
    for (int i = 0; i < size; ++i)
        minHeap->array[i] = newNode(data[i], freq[i]);
    minHeap->size = size;
    buildMinHeap(minHeap);
    return minHeap;
}

struct MinHeapNode *buildHuffmanTree(char data[], int freq[], int size) {
    struct MinHeapNode *left, *right, *top;
    struct MinHeap *minHeap = createAndBuildMinHeap(data, freq, size);
    while (!isSizeOne(minHeap)) {
        left = extractMin(minHeap);
        right = extractMin(minHeap);
        top = newNode('$', left->freq + right->freq);
        top->left = left;
        top->right = right;
        insertMinHeap(minHeap, top);
    }
    return extractMin(minHeap);
}

void printCodes(struct MinHeapNode *root, int arr[], int top) {
    if (root->left) {
        arr[top] = 0;
        printCodes(root->left, arr, top + 1);
    }
    if (root->right) {
        arr[top] = 1;
        printCodes(root->right, arr, top + 1);
    }
    if (isLeaf(root)) {
        printf("%c: ", root->data);
        printArr(arr, top);
    }
}

void HuffmanCodes(char data[], int freq[], int size) {
    struct MinHeapNode *root = buildHuffmanTree(data, freq, size);
    int arr[MAX_TREE_HT], top = 0;
    printCodes(root, arr, top);
}

int main() {
    char arr[] = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H'};
    int freq[] = {5, 25, 4, 7, 9, 12, 30, 8};
    int size = sizeof(arr) / sizeof(arr[0]);
    HuffmanCodes(arr, freq, size);
    return 0;
}

输出结果为：

A: 1101
B: 10
C: 11101
D: 11000
E: 1111
F: 0
G: 1110
H: 11001