用链表写的一个简单的进程 还有地方不知怎么实现 求指点啊

C/C++ code

#include <stdio.h>#include <stdlib.h> struct PCB{    char name[10];    int  rtime;    int  priority;    char state;    struct PCB *next;};/**初始化PCB*/struct PCB *initPCB(){  struct PCB *head;    head=(struct PCB *)malloc(sizeof(struct PCB));  if(NULL==head)  {      exit(-1);  }  head->next=NULL;  printf("please input the information of process\n");  printf("name:");  scanf("%s",head->name);  printf("run time:");  scanf("%d",&head->rtime);  printf("priority:");  scanf("%d",&head->priority);  head->state='R';  printf("\n");  return head;}/**找到优先级最高的进程*/void  MaxPriorityProcess(struct PCB * head,struct PCB* pCurrent){  struct PCB* temp;  struct PCB* p;  temp=head;  if(NULL==temp)//队列为空  {      temp=pCurrent;  }  if(pCurrent->priority>temp->priority)//如果优先级最大  {    pCurrent->next=temp;    temp=pCurrent;      }  if(NULL==temp->next)//队列里只有一个元素  {    if(temp->priority>pCurrent->priority)    {        temp->next=pCurrent;    }  }  /*队列元素排列*/  p=head;  while(p!=NULL)  {      if(temp->priority>pCurrent->priority)      {          p=temp;          temp=temp->next;      }      else      {          break;      }  }  pCurrent->next=p->next;  p->next=pCurrent;}/**打印PCB*/void printPCB(struct PCB * head){    struct PCB* temp;    temp=head->next;    while(temp!=NULL)    {        printf("\n process name: %s\n priority num: %d\n run time: %d\n process state:%c\n",            temp->name,temp->priority,temp->rtime,temp->state);        temp=temp->next;    }    printf("\n");}int main(){    struct PCB* head;    struct PCB* temp;    //head=new PCB;          PCB AA;    head = &AA;    for(int i=0;i<3;i++)              {      temp=initPCB();         temp->next = NULL;      MaxPriorityProcess(head,temp);               }    printPCB(head);                      return 0;}

处理器调度总是选队首进程运行。采用动态改变优先数的办法，进程每运行一次优先数就减“1”。由于本实习是模拟处理器调度，所以，对被选中的进程并不实际的启动运行，而是执行：
优先数-1
要求运行时间-1
来模拟进程的一次运行。
提醒注意的是：在实际的系统中，当一个进程被选中运行时，必须恢复进程的现场，让它占有处理器运行，直到出现等待事件或运行结束。在这里省去了这些工作。
(5) 进程运行一次后，若要求运行时间?0，则再将它加入队列（按优先数大小插入，且置队首标志）；若要求运行时间=0，则把它的状态修改成“结束”（E），且退出队列。

不知怎么实现

[解决办法]

C/C++ code

/**模拟简单的进程调度*/#include <stdio.h>#include <stdlib.h>typedef struct PCB{    char name[10];    int  rtime;    int  priorityNum;    char state;    struct PCB *next;}PCB;/**初始化PCB*/PCB *initPCB(){  PCB *head;  head=(PCB *)malloc(sizeof(PCB));  if(NULL==head)  {      exit(-1);  }  head->next=NULL;  return head;}/**  新建一个节点，并返回*/PCB *inputValue(){    PCB *temp = NULL;       temp = initPCB();    printf("please input the information of process\n");    printf("name:");    scanf("%s",temp->name);    printf("run time:");    scanf("%d",&temp->rtime);    printf("priorityNum:");    scanf("%d",&temp->priorityNum);    temp->state = 'R';    printf("\n");    return temp;}/**找到优先级最高的进程*/void findProcess( PCB * head, PCB * pCurrent ){       if( head->next == NULL )    {        head->next = pCurrent;        return ;    }       /* 从大到小排 */    PCB* p = head->next;    PCB* q = head;           while( p != NULL )    {        if( (pCurrent->priorityNum) > (p->priorityNum) )        {            q->next = pCurrent;            pCurrent->next = p;            break;           }        q = p;        p = p->next;    }    q->next = pCurrent;}/**打印PCB*/void printPCB( PCB * head ){    PCB *temp;    temp = head->next;    while( temp!=NULL )    {        printf("\n process name: %s run time: %d priority num: %d process state:%c",            temp->name,temp->rtime,temp->priorityNum,temp->state);        temp = temp->next;    }    printf("\n");}/**运行进程*/void runProcess( PCB * head){    PCB *processblock = NULL;    PCB *pcb = head->next;    while(pcb != NULL)    {           pcb->rtime -= 1;        pcb->priorityNum -= 1;        if(pcb->rtime == 0)        {            pcb->state = 'E';            pcb = pcb->next;            if(pcb == NULL)                return ;        }        else        {            processblock = pcb;            pcb = pcb->next;            processblock->next = NULL;            findProcess(pcb,processblock);            system("cls");        }        head->next = pcb;        printPCB(head);        getchar();    }    return ;}int main(){    PCB *head = NULL;    PCB *temp = NULL;    head = initPCB(); // 头结点为空    for(int i=0;i<5;i++)    {      temp = inputValue();      findProcess(head,temp);//其实是不符合要求的 它的复杂度为O(n) 只要和队首在比较一次 就可以了       printf("=================\n");           }   runProcess(head);   return 0;}