C++模拟ATM实现

下面代码实现的一个功能是模拟atm，程序允许用户输入3个数据：队列的最大长度、程序模拟的持续时间（单位为小时）以及平均每小时的客户数。

在程序中，使用循环，每次循环代表一分钟，在每次循环中，程序将完成下面的工作
1.是否来新客户，来：队列未满，入队；队列已满，拒绝
2.如果没有客户进行交易，则取队列的第一个客户。确定客户的已等候时间，并将wait_time计数器设置为新客户所需的处理时间
3.如果客户正在处理中，则讲wait_time计数器减1.
4.记录各种数据，如获得服务的客户数目、被拒绝的客户数（满队情况）、排队等候的累计时间以及累积的队列长度等。

queue.h

//queue.h --interface for a queue
#ifndef queue_h_
#define queue_h_

//this queue will contain customer items
class customer
{
private:
  long arrive; // arrive time for customer
  int processtime;//processing time for customer

public:
  customer() {arrive = processtime = 0;}

  void set(long when);
  long when() const {return arrive;}
  int ptime() const {return processtime; }
};

typedef customer item;

class queue
{
private:
  //class scope definitions
    //node is a nested structure definition local to this c
  struct node{item item; struct node *next;};
  enum {q_size = 10};

  //private class members
  node * front; // pointer to front of queue
  node * rear; // pointer to rear of queue
  int items; // current number of items in queue
  const int qsize; // maximum number of items in queue
  //preemptive definitions to prevent public copying
  queue(const queue & q) : qsize(0) {}
  queue & operator = (const queue & q) {return *this;}

public:
  queue(int qs = q_size); // create queue with a qs limit
  ~queue();
  bool isempty() const;
  bool isfull() const;
  int queuecount() const;
  bool enqueue(const item &item); // add item to end
  bool dequeue(item &item);
};

#endif

queue.cpp

//queue.cpp --queue and customer methods
#include 
#include "queue.h"

//queue methods
queue::queue(int qs) : qsize(qs)
{
  front = rear = null; //or nullptr 
  items = 0;
}

queue::~queue()
{
  node * temp;
  while (front != null) //while the queue is not yet empty
  {
    temp = front; // save the address of front item
    front = front->next;// reset pointer to next item
    delete temp; // delete former front
  }
}

bool queue::isempty() const
{
  return items == 0;
}

bool queue::isfull() const
{
  return items == qsize;
}

int queue::queuecount() const
{
  return items;
}

//add item to queue
bool queue::enqueue(const item& item)
{
  if(isfull())
    return false;
  node *add = new node; // create node
  // on failure, new throws std::bad_alloc exception
  add->item = item; // set node pointers
  add->next =null; // or nullptr
  items++;
  if(front == null) // if queue is empty
    front = add; // place item at front
  else
    rear->next = add; // else place at rear
  rear = add; // have rear point to new node
  return true;
}

//place front item into item variable and remove from queue
bool queue::dequeue(item& item)
{
  if(front == null)
      return false;
  item = front->item; // set item to first item in queue
  items--;
  node * temp = front; // save location of first item
  front = front->next; // rest front to next item
  delete temp; // delete former first item
  if(items == 0) 
    rear = null;
  return true;
}

//time set to a random value in  the range 1-3
void customer::set(long  when)
{
  processtime = std::rand() % 3 + 1;
  arrive = when;
}

main.cpp

//main.cpp -- using the queue interface
//compile with queue.cpp
#include 
#include  // for rand() and srand()
#include  // for time()
#include "queue.h"
const int min_per_hr = 60;

bool newcustomer(double x); // is there a new customer?

int main(int argc, char **argv) {
    using  std::cin;
    using  std::cout;
    using  std::endl;
    using  std::ios_base;

    //set things up
    std::srand(std::time(0)); //random initializing of rand()

    cout << " case study: bank of heather automatic teller \n";
    cout << "enter maximum size of queue: ";
    int qs;
    cin >> qs;
    queue line(qs); // line queue holds up to qs people

    cout << "enter the number of simulation hours: ";
    int hours; //hours of simulation
    cin >> hours;
    // simulation will run 1 cycle per minute
    long cyclelimit = min_per_hr * hours; // # of cycles

    cout << "enter the average number of customer per hour :  ";
    double perhour; //average # of arrival per hour
    cin >> perhour;
    double min_per_cust; //average time between arriavals
    min_per_cust = min_per_hr / perhour;

    item temp; //new customer data
    long turnaways = 0;// turned away by full queue
    long customers = 0;// joined the queue
    long served = 0;// served during the simulation
    long sum_line = 0;// cumulative line length
    int wait_time = 0;// time until autoteller is free
    long line_wait = 0;//cumulative time in line

    //running the simulation
    for(int cycle = 0; cycle < cyclelimit; cycle++)
    {
      if(newcustomer(min_per_cust))//have new customer
      {
    if(line.isfull())
      turnaways++;
    else{
      customers++;
      temp.set(cycle);//cycle = time of arrival
      line.enqueue(temp);// add newcomer to line
    }
      }
      if(wait_time <= 0 && !line.isempty())
      {
    line.dequeue(temp);//attend next customer
    wait_time = temp.ptime();//for wait_time minutes
    line_wait += cycle - temp.when();
    served++;
      }
      if(wait_time > 0)
    wait_time--;
      sum_line += line.queuecount();
    }

    //reporting results
    if(customers > 0)
    {
      cout << " customers accepted : " << customers << endl;
      cout << " customers served : " << served << endl;
      cout << "  turnaways : " << turnaways << endl;
      cout << " average queue size : ";
      cout.precision(2);
      cout << (double) sum_line / cyclelimit << endl;
      cout << " average wait time : " << (double) line_wait /served << " minutes\n";
    }
    else
      cout << " no customers!\n";
    cout << " done! \n ";

    return 0;
}

//x = average time in minutes, between customers
// return value is true if customer shoews up this minute
bool newcustomer(double x)
{
  return (std::rand() * x / rand_max < 1);
}

最后模拟的结果，如下图所示

可以看到如果一个小时客户的平均数量在15个时，拒绝的客户为0；等候时间也只有0.58min，但是如果将一个小时客户的平均数量设置在30个，等候的平均时间就会大大增加。这里因为采取随机数，所以每次结果可能会有一些不同。

同时，也练习了单链表的构建。链表是一种数据的存储方式，其保存的数据在内存中是不连续的，采用指针对数据进行访问，代码中这个写的非常清楚，而且也使用了很多次。

队列是一种数据结构，其特点是先进先出（正是因为如此，才特别符合atm的模拟）。