Question

更新：解决了我关于排序的第一个问题。但是现在我无法弄清楚如何在空闲时间内显示最早期限第一算法的正确图表。到目前为止，这是我的代码：

    import java.util.*;


    class deadlineprototype
    {


    public static void main(String args[])
    {


    Scanner sc = new Scanner(System.in);


    System.out.println("enter no. of processes : ");
    int n=sc.nextInt();
    int job[]=new int[n+1];
    int burst[]=new int[n+1];
    int newburst[]=new int[n+1];
    int arrival[]=new int[n+1];
    int deadline[]=new int[n+1];
    int wt[]=new int[n+1];
    int turn[]=new int[n+1];
    int tot_turn=0;
    int tot_wait=0;
    float avg_turn=0;
    float avg_wait=0;
    int j;

    for(int m=1;m<=n;m++)
        {
        arrival[m]=m;
        }
    for(int m=1;m<=n;m++)
        {
        job[m]=m;
        }

    for(int m=1;m<=n;m++)
        {
        System.out.println("enter arrival time, burst time and deadline of process "+(m)+"(0 for none):");
        arrival[m]=sc.nextInt();
        burst[m]=sc.nextInt();
        deadline[m]=sc.nextInt();

    if (deadline[m]==0){
        deadline[m]=1000;
    }
        }



    int temp;
    for(int i=0;i<n;i++)
{
   for(j=i;j<n;j++)
   {
      if(arrival[j+1] == arrival[j])
      { 
        if(deadline[j+1] < deadline[j])
        {
            temp=deadline[j+1];
            deadline[j+1]=deadline[j];
            deadline[j]=temp;

            temp=job[j+1];
            job[j+1]=job[j];
            job[j]=temp;

            temp=burst[j+1];
            burst[j+1]=burst[j];
            burst[j]=temp;


            temp=arrival[j+1];
            arrival[j+1]=arrival[j];
            arrival[j]=temp; 
        }
      }  
   }
}
    turn[1]=burst[1];    

    for(int i=2;i<=n;i++)
        {
        turn[i]=burst[i]+turn[i-1];
        wt[i]=turn[i]-burst[i];
        }
    for(int i=1;i<=n;i++)
        {
        tot_turn+=(wt[i]+burst[i])-arrival[i];
        avg_turn=(float)tot_turn/n;
        tot_wait+=wt[i]-arrival[i];
        avg_wait=(float)tot_wait/n;
        }
    System.out.println("----------Earliest Deadline Scheduling Diagram----------");
    for(int m=1;m<=n;m++)
        {
    if(deadline[m]==1000){
            deadline[m]=0;
            }
    if(wt[m]==0){
            System.out.println("0"+wt[m]+" _____"); 
    }
    else{
    System.out.println(wt[m]+" _____"); 
    }
            System.out.println("  |     |");

            System.out.println("  |job "+job[m]+"|");

            System.out.println("  |_____|");
                    try
                 {
                 //newburst[m]=(burst[m]*1000);
                 Thread.sleep(1000);  
                 }catch (InterruptedException ie)
                 {
                 System.out.println(ie.getMessage());
                 }
    }
    System.out.println((wt[wt.length-1]+burst[burst.length-1]));

如果我输入2个没有空闲的进程，那么它将显示正确的输出：

    enter no. of processes : 
    2
    enter arrival time, burst time and deadline of process 1(0 for none):
    0 17 0
    enter arrival time, burst time and deadline of process 2(0 for none):
    0 13 10
    ----------Earliest Deadline Scheduling Diagram----------
    00 _____
      |     |
      |job 2|
      |_____|
    13 _____
      |     |
      |job 1|
      |_____|
    30

但如果它有空闲时间，那么它将输出：

    enter no. of processes(5-10): 
    2
    enter arrival time, burst time and deadline of process 1(0 for none):
    0 5 0
    enter arrival time, burst time and deadline of process 2(0 for none):
    10 10 10
    ----------Earliest Deadline Scheduling Diagram----------
    00 _____
      |     |
      |job 1|
      |_____|
    5 _____
      |     |
      |job 2|
      |_____|
    15

我仍然坚持这样做，所以请帮助我。感谢

Answer 1

首先根据arrival time对进程进行排序，如下所示，

    for(i=0;i<n;i++)
    {
       for(j=i;j<n;j++)
       {
          if(arrival[j+1] < arrival[j])
          {
                temp=deadline[j+1];
                deadline[j+1]=deadline[j];
                deadline[j]=temp;

                temp=job[j+1];
                job[j+1]=job[j];
                job[j]=temp;

                temp=burst[j+1];
                burst[j+1]=burst[j];
                burst[j]=temp;


                temp=arrival[j+1];
                arrival[j+1]=arrival[j];
                arrival[j]=temp; 
          }
       }
    }

之后，如果arrival time进程相等，则根据deadline对其进行排序，如下所示，

    for(i=0;i<n;i++)
    {
       for(j=i;j<n;j++)
       {
          if(arrival[j+1] == arrival[j])
          { 
            if(deadline[j+1] < deadline[j])
            {
                temp=deadline[j+1];
                deadline[j+1]=deadline[j];
                deadline[j]=temp;

                temp=job[j+1];
                job[j+1]=job[j];
                job[j]=temp;

                temp=burst[j+1];
                burst[j+1]=burst[j];
                burst[j]=temp;


                temp=arrival[j+1];
                arrival[j+1]=arrival[j];
                arrival[j]=temp; 
            }
          }  
       }
    }

最早的截止日期cpu调度

1 个答案: