離散事件模擬

銀行業務的模擬程式

假設不存在插隊的情況，并且人人都會選擇排到人數最少的那一個視窗排隊，假設人們陸續到達銀行，不會同時到達銀行，

代碼如下：

#include<iostream>
#include<ctime>
#include<stdlib.h>
#define MAX_SIZE 10000//佇列的初始化長度
#define END_TIME 10000//銀行關門時間
using namespace std;
typedef struct TIME {//
	int daoda_time;//客戶到達時間
	int likai_time;//客戶離開時間
	int banli_time;//客戶辦理業務所需時間
}time1;
typedef struct Windows {
	TIME* point;
	int front;//記入佇列的隊頭位置
	int rear;//即入佇列的隊尾位置
	int size;//定義的線表可以使用的長度
	int length;//已經使用的長度
}window;
window list1, list2, list3, list4;
int n1, n2, n3, n4;//計時器
int nn;//計時器
int all_time;
int people_num;
int record1;//計入亂數，記錄當前到達的客戶辦理業務所需的時間
int record2;//記入亂數，記入下一客戶到達的間隔時間
void moni_daoda(int e);
int search_min();
void banli_go(window* a, int b, int& c);
int main()
{
	srand(time(0));//設定亂數種子
	record1 = rand() % 5 + 5;
	record2 = rand() % 3 + 1;
	cout << record1 << " " << record2 << endl;
	list1.point = (TIME*)malloc(MAX_SIZE * sizeof(TIME));//初始化四個視窗佇列
	list1.size = MAX_SIZE; list1.front = 0;
	list1.point[0].daoda_time = 0; list1.point[0].banli_time = record1;
	list1.rear = 1; list1.length = 1;
	people_num = 1;
	list2.point = (TIME*)malloc(MAX_SIZE * sizeof(TIME));
	list2.front = 0; list2.length = 0; list2.rear = 0; list2.size = MAX_SIZE;
	list3.point = (TIME*)malloc(MAX_SIZE * sizeof(TIME));
	list3.front = 0; list3.length = 0; list3.rear = 0; list3.size = MAX_SIZE;
	list4.point = (TIME*)malloc(MAX_SIZE * sizeof(TIME));
	list4.front = 0; list4.length = 0; list4.rear = 0; list4.size = MAX_SIZE;
	for (int n = 0; n <= END_TIME; n++)
	{
		moni_daoda(n);
		if (list1.length != 0) banli_go(&list1, n, n1);
		if (list2.length != 0) banli_go(&list2, n, n2);
		if (list3.length != 0) banli_go(&list3, n, n3);
		if (list4.length != 0) banli_go(&list4, n, n4);
	}

	cout << "平均每個客戶在銀行的逗留時間為：" << all_time / people_num << endl;
	return 0;
}
void moni_daoda(int e)模擬客戶到達銀行，并且尋找視窗排隊，
{
	int i;
	if (nn < record2)
	{
		nn++;
	}
	else//到達間隔時間后新的客戶到達
	{
		record1 = rand() % 5 + 5;
		record2 = rand() % 3 + 1;
		i = search_min();//尋找人數最小的哪一個視窗
		if (i == 1)//客戶到達后的選視窗入隊操作
		{
			list1.point[list1.rear].daoda_time = e; list1.point[list1.rear].banli_time = record1;
			list1.length = list1.length + 1; list1.rear = list1.rear + 1;
			people_num++;
		}
		else if (i == 2)
		{
			list2.point[list2.rear].daoda_time = e; list2.point[list2.rear].banli_time = record1;
			list2.length = list2.length + 1; list2.rear = list2.rear + 1;
			people_num++;
		}
		else if (i == 3)
		{
			list3.point[list3.rear].daoda_time = e; list3.point[list3.rear].banli_time = record1;
			list3.length = list3.length + 1; list3.rear = list3.rear + 1;
			people_num++;
		}
		else
		{
			list4.point[list4.rear].daoda_time = e; list4.point[list4.rear].banli_time = record1;
			list4.length = list4.length + 1; list4.rear = list4.rear + 1;
			people_num++;
		}
		nn = 0;
	}
}
int search_min()//尋找人數最小的那一個視窗的函式
{
	if (list1.length >= list2.length || list1.length >= list3.length || list1.length >= list4.length) return 1;
	else if (list2.length > list1.length || list2.length >= list3.length || list2.length >= list4.length) return 2;
	else if (list3.length > list1.length || list3.length > list2.length || list3.length >= list4.length) return 3;
	else return 4;
}
void banli_go(window* a, int b, int& c)//客戶辦理好業務后的出隊操作
{
	if (c < a->point[a->front].banli_time)
	{
		c++;
	}
	else//出隊操作
	{
		a->point[a->front].likai_time = b;
		all_time = all_time + a->point[a->front].likai_time - a->point[a->front].daoda_time;
		a->length = a->length - 1;
		a->front = a->front + 1;
		c = 0;
	}
}