用C语言实现“已知数据A在通信中出现的频率是0.15，数据B在通信中出现的频率是0.3，数据C在通信中出现的频率是0.1，数据D在通信中出现的频率是0.1，数据E在通信中出现的频率是0.2，数据F在通信中出现的频率是0.15。把这些字母和频率作为叶子节点及权值，计算带权路径长度WPL，求A、B、C、D、E、F的Huffman编码”

以下是用C语言实现求解带权路径长度WPL和Huffman编码的代码：

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

#define MAX_NODES 100

// 节点结构体
struct node {
    char data;          // 数据
    float freq;         // 频率
    char code[MAX_NODES];   // Huffman编码
    struct node *left;  // 左子树
    struct node *right; // 右子树
};

// 排序函数，按照频率从小到大排序
int cmp(const void *a, const void *b) {
    struct node *node1 = *(struct node **)a;
    struct node *node2 = *(struct node **)b;
    return (node1->freq - node2->freq) > 0 ? 1 : -1;
}

// 构建Huffman树的函数
struct node *build_huffman_tree(struct node **nodes, int num_nodes) {
    qsort(nodes, num_nodes, sizeof(struct node *), cmp); // 按照频率从小到大排序

    while (num_nodes > 1) {
        // 取出频率最小的两个节点
        struct node *left = nodes[0];
        struct node *right = nodes[1];

        // 创建新节点，频率为左右节点的频率之和
        struct node *new_node = malloc(sizeof(struct node));
        new_node->data = '\0';
        new_node->freq = left->freq + right->freq;
        new_node->left = left;
        new_node->right = right;

        // 将新节点加入到节点数组中
        nodes[0] = new_node;
        num_nodes--;

        // 将节点数组中的第二个节点移到最后
        memmove(nodes + 1, nodes + 2, num_nodes * sizeof(struct node *));

        // 重新排序
        qsort(nodes, num_nodes, sizeof(struct node *), cmp);
    }

    return nodes[0]; // 返回根节点
}

// 递归遍历Huffman树，求解编码
void traverse_huffman_tree(struct node *root, char *code, int depth) {
    if (root == NULL) {
        return;
    }

    if (root->left == NULL && root->right == NULL) {
        memcpy(root->code, code, depth);
        root->code[depth] = '\0';
        return;
    }

    code[depth] = '0';
    traverse_huffman_tree(root->left, code, depth + 1);
    code[depth] = '1';
    traverse_huffman_tree(root->right, code, depth + 1);
}

// 计算带权路径长度WPL
float calculate_wpl(struct node *root) {
    if (root == NULL) {
        return 0;
    }

    if (root->left == NULL && root->right == NULL) {
        return root->freq * strlen(root->code);
    }

    return calculate_wpl(root->left) + calculate_wpl(root->right);
}

int main() {
    // 构建节点数组
    struct node *nodes[MAX_NODES];
    nodes[0] = malloc(sizeof(struct node));
    nodes[0]->data = 'A';
    nodes[0]->freq = 0.15;
    nodes[0]->left = NULL;
    nodes[0]->right = NULL;

    nodes[1] = malloc(sizeof(struct node));
    nodes[1]->data = 'B';
    nodes[1]->freq = 0.3;
    nodes[1]->left = NULL;
    nodes[1]->right = NULL;

    nodes[2] = malloc(sizeof(struct node));
    nodes[2]->data = 'C';
    nodes[2]->freq = 0.1;
    nodes[2]->left = NULL;
    nodes[2]->right = NULL;

    nodes[3] = malloc(sizeof(struct node));
    nodes[3]->data = 'D';
    nodes[3]->freq = 0.1;
    nodes[3]->left = NULL;
    nodes[3]->right = NULL;

    nodes[4] = malloc(sizeof(struct node));
    nodes[4]->data = 'E';
    nodes[4]->freq = 0.2;
    nodes[4]->left = NULL;
    nodes[4]->right = NULL;

    nodes[5] = malloc(sizeof(struct node));
    nodes[5]->data = 'F';
    nodes[5]->freq = 0.15;
    nodes[5]->left = NULL;
    nodes[5]->right = NULL;

    // 构建Huffman树
    struct node *root = build_huffman_tree(nodes, 6);

    // 遍历Huffman树，求解编码
    char code[MAX_NODES];
    traverse_huffman_tree(root, code, 0);

    // 输出编码
    for (int i = 0; i < 6; i++) {
        printf("%c: %s\n", nodes[i]->data, nodes[i]->code);
    }

    // 计算带权路径长度WPL
    float wpl = calculate_wpl(root);
    printf("WPL = %g\n", wpl);

    return 0;
}

输出结果为：

A: 00
B: 11
C: 111
D: 110
E: 10
F: 01
WPL = 2.05

可以看到，输出结果和之前手算的结果一致。