在循环调度中查找给定 N 个进程的执行顺序

给定一个数组arr[] ，表示使用循环算法调度的N 个进程的突发时间，并具有给定的量子时间Q 。假设所有进程在时间t = 0到达，任务是找到进程执行的顺序。

例子：

Input: arr[] = {3, 7, 4}, q = 3
Output: 0 2 1
Explanation:
The order of execution is as follows P0, P1, P2, P1, P2, P1
Since, P0 has burst time of 3 and quantum time is also 3, it gets completed first.
P1 has burst time of 7 so after executing for 3 units, it gets context switched and P2 executes.
P2 has burst time of 4 so after executing for 3 units, it gets context switched and P1 executes.
Again P1 starts executing since it has 4 units burst time left, so it executes for another 3 units and then context switches.
Now process P2 executes for 1 unit and gets completed.
In the end process P1 is completed.
They complete the execution in the order P0, P2, P1.

Input: arr[] = {13, 8, 5}, q = 6
Output: 2 1 0
Explanation: 
Initially P0 starts and after 6 units, its context switches.
P1 executes for 6 units and context switches.
Since P2 has burst time less than quantum time, so it executes for 5 units and gets completed first.
P0 has remaining burst time 7 units, so it executes again for 6 units and context switches.
P1 has remaining burst time as 2 units and it gets completed second.
In the end process P0 gets completed.
They complete the execution in the order P2, P1, P0.

方法：这个想法是创建一个辅助数组，其中包含进程与 CPU 交互次数的频率。然后freq[]数组可以按频率的递增顺序排序。首先完成频率最低的过程，然后是第二个过程，依此类推。排序后的数组给出了过程的完成顺序。以下是步骤：

• 初始化一个数组freq[] ，其中freq[i]是第i^个进程与 CPU 交互的次数。
• 初始化数组order[]以存储进程的完成顺序并存储order[i] = i 。
• 按照freq[]的递增顺序对数组order[]进行排序，使得freq[order[i]] ≤ freq[order[i + 1]] 。
• 打印数组order[] ，这是进程完成执行的顺序。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to sort the array order[]
// on the basis of the array freq[]
void merge(int* order, int* freq, int i,
           int mid, int j)
{
    int tempOrder[j - i + 1];
    int temp = mid + 1, index = -1;
 
    while (i <= mid && temp <= j) {
 
        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]]
            <= freq[order[temp]]) {
            tempOrder[++index] = order[i++];
        }
 
        // Otherwise
        else {
            tempOrder[++index] = order[temp++];
        }
    }
 
    // Add the left half to tempOrder[]
    while (i <= mid) {
        tempOrder[++index] = order[i++];
    }
 
    // Add right half to tempOrder[]
    while (temp <= j) {
        tempOrder[++index] = order[temp++];
    }
 
    // Copy the tempOrder[] array to
    // order[] array
    for (index; index >= 0; index--) {
        order[j--] = tempOrder[index];
    }
}
 
// Utility function to sort the array
// order[] on the basis of freq[]
void divide(int* order, int* freq,
            int i, int j)
{
    // Base Case
    if (i >= j)
        return;
 
    // Divide array into 2 parts
    int mid = i / 2 + j / 2;
 
    // Sort the left array
    divide(order, freq, i, mid);
 
    // Sort the right array
    divide(order, freq, mid + 1, j);
 
    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}
 
// Function to find the order of
// processes in which execution occurs
void orderProcesses(int A[], int N, int q)
{
    int i = 0;
 
    // Store the frequency
    int freq[N];
 
    // Find elements in array freq[]
    for (i = 0; i < N; i++) {
        freq[i] = (A[i] / q)
                  + (A[i] % q > 0);
    }
 
    // Store the order of completion
    // of processes
    int order[4];
 
    // Initialize order[i] as i
    for (i = 0; i < N; i++) {
        order[i] = i;
    }
 
    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);
 
    // Print order of completion
    // of processes
    for (i = 0; i < N; i++) {
        cout << order[i] << "  ";
    }
}
 
// Driver Code
int main()
{
    // Burst Time of the processes
    int arr[] = { 3, 7, 4 };
 
    // Quantam Time
    int Q = 3;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    orderProcesses(arr, N, Q);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to sort the array order[]
// on the basis of the array freq[]
static void merge(int order[], int freq[], int i,
           int mid, int j)
{
    int tempOrder[] = new int[j - i + 1];
    int temp = mid + 1, index = -1;
 
    while (i <= mid && temp <= j)
    {
 
        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]]
            <= freq[order[temp]])
        {
            tempOrder[++index] = order[i++];
        }
 
        // Otherwise
        else
        {
            tempOrder[++index] = order[temp++];
        }
    }
 
    // Add the left half to tempOrder[]
    while (i <= mid)
    {
        tempOrder[++index] = order[i++];
    }
 
    // Add right half to tempOrder[]
    while (temp <= j)
    {
        tempOrder[++index] = order[temp++];
    }
 
    // Copy the tempOrder[] array to
    // order[] array
  int ind= index;
    for (index= ind; index >= 0; index--)
    {
        order[j--] = tempOrder[index];
    }
}
 
// Utility function to sort the array
// order[] on the basis of freq[]
static void divide(int order[], int freq[],
            int i, int j)
{
    // Base Case
    if (i >= j)
        return;
 
    // Divide array into 2 parts
    int mid = i / 2 + j / 2;
 
    // Sort the left array
    divide(order, freq, i, mid);
 
    // Sort the right array
    divide(order, freq, mid + 1, j);
 
    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}
 
// Function to find the order of
// processes in which execution occurs
static void orderProcesses(int A[], int N, int q)
{
    int i = 0;
 
    // Store the frequency
    int freq[] = new int[N];
 
    // Find elements in array freq[]
    for (i = 0; i < N; i++)
    {
        freq[i] = (A[i] / q);
        if (A[i] % q > 0)
              freq[i] += 1;
    }
 
    // Store the order of completion
    // of processes
    int order[] = new int[4];
 
    // Initialize order[i] as i
    for (i = 0; i < N; i++) {
        order[i] = i;
    }
 
    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);
 
    // Print order of completion
    // of processes
    for (i = 0; i < N; i++) {
        System.out.print( order[i] + "  ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
   
    // Burst Time of the processes
    int arr[] = { 3, 7, 4 };
 
    // Quantam Time
    int Q = 3;
 
    int N = arr.length;
 
    // Function Call
    orderProcesses(arr, N, Q);
}
}
 
// This code is contributed by chitranayal.

C#

// C# program for the above approach
using System;
 
class GFG{
  
// Function to sort the array order[]
// on the basis of the array freq[]
static void merge(int[] order, int[] freq, int i,
                  int mid, int j)
{
    int[] tempOrder = new int[j - i + 1];
    int temp = mid + 1, index = -1;
  
    while (i <= mid && temp <= j)
    {
         
        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]] <= freq[order[temp]])
        {
            tempOrder[++index] = order[i++];
        }
  
        // Otherwise
        else
        {
            tempOrder[++index] = order[temp++];
        }
    }
  
    // Add the left half to tempOrder[]
    while (i <= mid)
    {
        tempOrder[++index] = order[i++];
    }
  
    // Add right half to tempOrder[]
    while (temp <= j)
    {
        tempOrder[++index] = order[temp++];
    }
  
    // Copy the tempOrder[] array to
    // order[] array
    int ind = index;
    for(index = ind; index >= 0; index--)
    {
        order[j--] = tempOrder[index];
    }
}
  
// Utility function to sort the array
// order[] on the basis of freq[]
static void divide(int[] order, int[] freq,
                   int i, int j)
{
     
    // Base Case
    if (i >= j)
        return;
  
    // Divide array into 2 parts
    int mid = i / 2 + j / 2;
  
    // Sort the left array
    divide(order, freq, i, mid);
  
    // Sort the right array
    divide(order, freq, mid + 1, j);
  
    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}
  
// Function to find the order of
// processes in which execution occurs
static void orderProcesses(int[] A, int N, int q)
{
    int i = 0;
     
    // Store the frequency
    int[] freq = new int[N];
  
    // Find elements in array freq[]
    for(i = 0; i < N; i++)
    {
        freq[i] = (A[i] / q);
         
        if (A[i] % q > 0)
              freq[i] += 1;
    }
  
    // Store the order of completion
    // of processes
    int[] order = new int[4];
  
    // Initialize order[i] as i
    for(i = 0; i < N; i++)
    {
        order[i] = i;
    }
  
    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);
  
    // Print order of completion
    // of processes
    for(i = 0; i < N; i++)
    {
        Console.Write( order[i] + "  ");
    }
}
  
// Driver Code
public static void Main()
{
     
    // Burst Time of the processes
    int[] arr = { 3, 7, 4 };
  
    // Quantam Time
    int Q = 3;
  
    int N = arr.Length;
  
    // Function Call
    orderProcesses(arr, N, Q);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// JavaScript program for the above approach
 
// Function to sort the array order[]
// on the basis of the array freq[]
function merge(order, freq, i, mid, j)
{
    var tempOrder = Array(j - i + 1);
    var temp = mid + 1, index = -1;
 
    while (i <= mid && temp <= j) {
 
        // If order[i]th is less than
        // order[temp]th process
        if (freq[order[i]]
            <= freq[order[temp]]) {
            tempOrder[++index] = order[i++];
        }
 
        // Otherwise
        else {
            tempOrder[++index] = order[temp++];
        }
    }
 
    // Add the left half to tempOrder[]
    while (i <= mid) {
        tempOrder[++index] = order[i++];
    }
 
    // Add right half to tempOrder[]
    while (temp <= j) {
        tempOrder[++index] = order[temp++];
    }
 
    // Copy the tempOrder[] array to
    // order[] array
    for (index; index >= 0; index--) {
        order[j--] = tempOrder[index];
    }
}
 
// Utility function to sort the array
// order[] on the basis of freq[]
function divide(order, freq, i, j)
{
    // Base Case
    if (i >= j)
        return;
 
    // Divide array into 2 parts
    var mid = parseInt(i / 2) + parseInt(j / 2);
 
    // Sort the left array
    divide(order, freq, i, mid);
 
    // Sort the right array
    divide(order, freq, mid + 1, j);
 
    // Merge the sorted arrays
    merge(order, freq, i, mid, j);
}
 
// Function to find the order of
// processes in which execution occurs
function orderProcesses(A, N, q)
{
    var i = 0;
 
    // Store the frequency
    var freq = Array(N);
 
    // Find elements in array freq[]
    for (i = 0; i < N; i++) {
        freq[i] = (A[i] / q)
                  + (A[i] % q > 0);
    }
 
    // Store the order of completion
    // of processes
    var order = Array(4);
 
    // Initialize order[i] as i
    for (i = 0; i < N; i++) {
        order[i] = i;
    }
 
    // Function Call to find the order
    // of execution
    divide(order, freq, 0, N - 1);
 
    // Print order of completion
    // of processes
    for (i = 0; i < N; i++) {
        document.write( order[i] + "  ");
    }
}
 
// Driver Code
 
// Burst Time of the processes
var arr = [3, 7, 4];
 
// Quantam Time
var Q = 3;
var N = arr.length;
 
// Function Call
orderProcesses(arr, N, Q);
 

输出：

0  2  1

时间复杂度： O(N*log N)
辅助空间： O(N)

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