Simulieren einer Kundenlinie mit verknüpften Liste in C++

Question

Ich versuche, eine Kundenlinie zu simulieren, die Kundennummer, Wartezeiten usw. zeigen. Ich habe drei Benutzereingaben: Max Simulationszeit, max Intervall zwischen Kundeneinträgen und max Servicezeit.

schließlich, würde Ich mag eine Ausgabe erzeugen, die wie so aussieht:

Line simulation output

hier ist mein Code sofar, aber im nur in der Lage, die Minuten zu generieren, aber nicht den Rest des Ausgangs.

#include "stdafx.h" 
#include <iostream> 
using namespace std; 


class Node { 
public: 
    Node() {}; 
    Node(int, int, int) {}; 
    //Node(int,int, int){} 
    void SetCustEntryMinute(int custNum, int entryMinute, int servTimeRem) { 

     entryMin = entryMinute; 
     custNo = custNum; 
     serviceTimeRemain = servTimeRem; 
     next = NULL; 
    } 
    void SetNext(Node* aNext) { next = aNext; } 
    int CustEntryMinute() { return entryMin; } 
    int CustNum() { return custNo; } 
    int ServTmRm() { return serviceTimeRemain; } 
    Node* Next() { return next; } 

    int custNo; 
    int entryMin; 
    int serviceTimeRemain; 
    Node* next; 
}; 


//List class 
class List { 
public: 
    List() { /*head = NULL; tail = NULL; */}; 
    //void Print(); 
    void enqueue(int custNum, int entryMinute, int servTimeRem); 
    void dequeue(); 
    Node *getHead() { return head; }; 
    Node* head; 
    Node *tail; 
}; 



/* Append a node to the linked list*/ 
void List::enqueue(int custNum, int entryMinute, int servTimeRem) 
{ 

    // Create a new node 
    Node* newNode = new Node(custNum, entryMinute, servTimeRem); 

    // Create a temp pointer 
    if (tail) 
    { 
     tail->next = newNode; 
    } 
    else 
    { 
     head = newNode; 
    } 
    tail = newNode; 

} 



/* Serve a node from the list*/ 
void List::dequeue() { 

    // Create a temp pointer 
    Node *tmp = head; 
    //head = head->Next(); 

    // No nodes 
    if (tmp == NULL) 
    { 
     //nothing to delete 
     return; 
    } 
    if (head == tail)//only one node 
    { 
     head = NULL; 
     tail = NULL; 
    } 
    else 
    { 
     head = head->next; 
    } 
    delete tmp; 
} 



int main() 
{ 
    List list; 
    int serviceTimeMax; 
    int maxInterval; 
    int timeCount = 0; 
    int timeMax; 
    int entryMinute = 0; 

    cout << "Enter max time: "; 
    cin >> timeMax; 

    cout << "Enter max interval between two services: "; 
    cin >> maxInterval; 

    cout << "Enter max service time: "; 
    cin >> serviceTimeMax; 

    int custNum = 1; 
    Node* personAhead = NULL; 
    cout << "Minute \tCustomer\t Entry\t Service Time\t Wait time\t Minutes remaining\n"; 
    cout << "Number \tNumber \tMinute\t Remaining\t Remaining\t until service is completed\n\n"; 
    int nextJobTime = 1; 
    for (int i = 1; i < timeMax + 1; i++) { 
     if (i == nextJobTime) 
     { 
      list.enqueue(custNum, i, rand() % serviceTimeMax); 
       nextJobTime = i + rand() % maxInterval; 
     } 
     Node *head = new Node; 
     while (head != NULL) { 
      if (head->serviceTimeRemain == 0) 
      { 
       list.dequeue(); 
       head = list.getHead(); 
      } 

      Node *p = head; 
      int wait_time = 0; 
      cout << i << "\n"; 
      while (p != NULL) 
      { 
       wait_time += p->serviceTimeRemain; 
       cout << "\t" << p->custNo << "\t" << p->entryMin << "\t" << wait_time << "\t" << wait_time + p->serviceTimeRemain << "\n"; 
       p->serviceTimeRemain--; 
      } 
     } 


    } 


    system("pause"); 
    return 0; 
} 

Zu diesem Zeitpunkt wird nur die vom Benutzer eingegebene Zeit generiert.

Answer 1

Sie sind in einer Endlos-Schleife

while (p != NULL) 
{ 
    wait_time += p->serviceTimeRemain; 
    cout << "\t" << p->custNo << "\t" << p->entryMin << "\t" << wait_time << "\t" << wait_time + p->serviceTimeRemain << "\n"; 
    p->serviceTimeRemain--; 
} 

An keiner Stelle stecken Sie ändern Sie p, was bedeutet, werden Sie nie die while-Schleife verlassen.