c语言数据结构的舞会问题

Question

问题：对于舞伴配对问题，先人队的男士或女士先出队配成舞伴，因此设置两个队列分别存放男士和女士人队者。假设男士和女士的记录存放在一个数组中作为输人，然后依次扫描该数组的各元素，并根据性别来决定是进入男队还是女队。当这两个队列构造完成之后，依次将两队当前的队头元素出队来配成舞伴，直至某队列变空为止。此时，若某队仍有等待配对者，则输出此队列中排在队头的等待者的姓名，此人将是下一轮舞曲开始时第一个可获得舞伴的人。
已经写了一部分，希望能办我修改和补充一下代码，以完善上面的功能

 
#include<stdio.h>
#include<stdlib.h>
 
type struct
{
    char name[20];
    char sex;
}Person;
 
#define MAXQSIZE 100
typedef struct
{
    Person *base;
    int front;
    int rear;
}SqQueue;
SqQueue Mdancers,Fdancers;
 
Status InitQueue(SqQueue &Q)
{
    Q.base=new QElemType[MAXQSIZE];
    if(!Q.base)
        exit (OVERFLOW);
    Q.front =Q.rear=0;
    return OK
}
 
void DancePartner(Person dancer[],int num)
{
    InitQueue(Mdancers);
    InitQueue(Fdancers);
    for(i=0;i<num;i++)
    {
        p=dancer[i];
        if(p.sex=='F')
             EnQueue(Fdancer,p);
        else
             EnQueue(Mdancers,p);
    }
    cout<<"the dancing partners are:\n;
    while(!QueueEmpty(Fdancers)&&!QueueEmpty(Mdancers))
    {
        DeQueue(Fdancers,p);
        cout<<p.name<<" ";
        DeQueue(Mdancers,p);
        cout<<p.name<<endl;
    }
    if(!QueueEmpty(Fdancers));
    {
        p=GetHead(Mdancers)
        cout<<"the first man to get a partner is:"<<p.name<<endl;
    }
}
int main()
{
   
}

Answer 1

你这代码C和C++混杂，代码修改如下：
运行结果：

代码：

#define _CRT_SECURE_NO_WARNINGS


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

#define MAXQSIZE 100
typedef int Status;
#define OK 1
#define OVERFLOW -1



typedef struct
{
    char name[20];
    char sex;
}Person;


typedef struct
{
    Person* base;
    int front;
    int rear;
}SqQueue;

SqQueue Mdancers, Fdancers;

Status InitQueue(SqQueue* Q)
{
    Q->base = (Person*)malloc(sizeof(Person) * MAXQSIZE);
    if (!Q->base)
        return OVERFLOW;
    Q->front = Q->rear = 0;
    return OK;
}
//入队
Status EnQueue(SqQueue* Q, Person p)
{
    if (Q->rear == MAXQSIZE - 1)
        return OVERFLOW;
    Q->base[Q->rear] = p;
    Q->rear++;
    return OK;
}
//出队
Status DeQueue(SqQueue* Q, Person* p)
{
    if (Q->front == Q->rear)
        return OVERFLOW;
    *p = Q->base[Q->front];
    Q->front++;
    return OK;
}

//判断队列是否为空
int QueueEmpty(SqQueue Q)
{
    if (Q.front == Q.rear)
        return 1;
    else
        return 0;
}

//获取队列的第一个元素
Person GetTop(SqQueue Q)
{
    Person p = Q.base[Q.front];
    return p;
}


void DancePartner(Person dancer[], int num)
{
    int i;
    Person p;
    InitQueue(&Mdancers);
    InitQueue(&Fdancers);
    for (i = 0; i < num; i++)
    {
        p = dancer[i];
        if (p.sex == 'F')
            EnQueue(&Fdancers, p);
        else
            EnQueue(&Mdancers, p);
    }
    printf("the dancing partners are:\n");
    while (!QueueEmpty(Fdancers) && !QueueEmpty(Mdancers))
    {
        DeQueue(&Fdancers, &p);
        printf("%s ", p.name);
        DeQueue(&Mdancers, &p);
        printf("%s\n", p.name);
    }
    if (!QueueEmpty(Fdancers))
    {
        p = GetTop(Fdancers);
        printf("the first woman to get a partner is:%s\n", p.name);
    }
    if (!QueueEmpty(Mdancers))
    {
        p = GetTop(Mdancers);
        printf("the first man to get a partner is:%s\n", p.name);
    }
}
int main()
{
    Person ps[] = { {"ZhangSan",'M'},{"LiLi",'F'},{"ZhaoQian",'F'},{"WangJuan",'F'},{"WangGang",'M'} };
    DancePartner(ps, 5);
    return 0;
}