数据结构中的舞伴问题

Question

假设在周末舞会上，男士们和女士们在进入舞厅时，各自排成一队。跳舞开始时，依次
从男队和女队的队头各出一人配成舞伴。若两队初始人数不相同，则较长的那一队中未配对
者等待下一轮舞曲。一支舞曲结束所有刚跳过舞的人按原来的顺序排到原队列的后面。请输
出前三支舞曲的男女配对名单。

Answer 1

我的思路是使用队列这种数据结构，先入先出，每个人跳完舞之后再次入队，等待下一轮舞曲。
使用链表构造的一个队列，代码如下，代码哪里不对或者不好，请告诉我。

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

#define QUEUE_MAX_LENGHT 20

struct Element
{
    char name[20];
    struct Element *pre;
    struct Element *next;
};

struct Queue
{
    int length;
    struct Element *head;
    struct Element *tail;
};

// 出队
struct Element *take(struct Queue *queue)
{
    if (queue->length < 1)
    {
        printf("Queue is empty.");
        exit(1);
    }
    struct Element *old_head = queue->head;

    struct Element *new_head = old_head->next;
    queue->head = new_head;
    queue->length--;

    old_head->next = NULL;
    old_head->pre = NULL;
    return old_head;
}

// 入队
void push(struct Queue *queue, struct Element *element)
{
    if (queue->length >= 20)
    {
        printf("Queue too long.");
        exit(1);
    }

    if (queue->tail == NULL)
    {
        queue->head = element;
        queue->tail = element;
        queue->length++;
    }
    else
    {
        struct Element *old_tail = queue->tail;
        old_tail->next = element;
        element->pre = old_tail;

        queue->tail = element;
        queue->length++;
    }
}

struct Queue *init_queue()
{
    struct Queue *queue = (struct Queue *)malloc(sizeof(struct Queue));
    queue->length = 0;
    queue->head = NULL;
    queue->tail = NULL;
    return queue;
}

struct Element *init_element(char *name)
{
    struct Element *element = (struct Element *)malloc(sizeof(struct Element));
    strcpy(element->name, name);
    element->pre = NULL;
    element->next = NULL;
    return element;
}

int main()
{
    struct Queue *man_queue = init_queue();
    struct Queue *miss_queue = init_queue();

    int i;
    // 初始化男士队列
    for (i = 0; i < 10; i++)
    {
        char name[20] = "man ";

        char num[10];
        sprintf(num, "%d", i);

        strcat(name, num);

        struct Element *new_man = init_element(name);
        push(man_queue, new_man);
    }

    // 初始化女士队列
    for (i = 0; i < 8; i++)
    {
        char name[20] = "miss ";

        char num[10];
        sprintf(num, "%d", i);

        strcat(name, num);

        struct Element *new_miss = init_element(name);
        push(miss_queue, new_miss);
    }

    int min_num;
    min_num = (miss_queue->length > man_queue->length) ? man_queue->length : miss_queue->length; // 取短队伍的数量。

    int j;
    for (i = 0; i < 3; i++) // 三支舞曲
    {
        printf("dance music %d\n", i + 1);

        for (j = 0; j < min_num; j++) // 短队伍的人都要匹配到。
        {
            struct Element *miss = take(miss_queue); // 取出队列中排队的第一人。
            struct Element *man = take(man_queue);
            printf("dance partners is '%s' and '%s'\n", miss->name, man->name);
            // 跳舞中
            push(miss_queue, miss); // 跳完舞之后再次入队，等待下一首舞曲。
            push(man_queue, man);
        }
    }
}