1017 Queueing at Bank (25 分)

最新推荐文章于 2020-05-24 23:57:17 发布

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本文介绍了一个银行服务系统的模拟算法，该算法通过计算不同窗口的结束服务时间来预测客户的平均等待时间。考虑到银行的营业时间限制，任何在17点之后到达的客户将不被服务。通过对客户到达时间和处理时间的输入进行排序，算法有效地分配了服务窗口，并计算了所有客户的平均等待时间。

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Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and there is a window available. It is assumed that no window can be occupied by a single customer for more than 1 hour.

Now given the arriving time T and the processing time P of each customer, you are supposed to tell the average waiting time of all the customers.

Input Specification:
Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤10
4
) - the total number of customers, and K (≤100) - the number of windows. Then N lines follow, each contains 2 times: HH:MM:SS - the arriving time, and P - the processing time in minutes of a customer. Here HH is in the range [00, 23], MM and SS are both in [00, 59]. It is assumed that no two customers arrives at the same time.

Notice that the bank opens from 08:00 to 17:00. Anyone arrives early will have to wait in line till 08:00, and anyone comes too late (at or after 17:00:01) will not be served nor counted into the average.

Output Specification:
For each test case, print in one line the average waiting time of all the customers, in minutes and accurate up to 1 decimal place.

Sample Input:
7 3
07:55:00 16
17:00:01 2
07:59:59 15
08:01:00 60
08:00:00 30
08:00:02 2
08:03:00 10
Sample Output:
8.2

相比于1014，这道题似乎要简单一点，相当于是黄线里面只有一排。
注意：只要顾客在17点前到达，那么一定会被服务。
最后一个测试点中的LastFinishTime 可能会大于24点，所以在找最小的窗口时应该将MinTime初始化更大。
或者直接用LastFinishTime[0]为初值，同时从1开始循环，找最小的时间。
在这里插入图片描述

#include <iostream>
#include <vector>
#include<algorithm>

using namespace std;
struct CustomerData{
	int ArriveTime;
	int ProcessTime;
	int WaitingTime = 0;
};

bool cmp(CustomerData A, CustomerData B)
{
	return A.ArriveTime < B.ArriveTime;
}

int main(void)
{
	freopen("input2.txt", "r", stdin);
	int CustomerNum, WindowNum;
	int Hour, Minute, Second, Time;
	cin >> CustomerNum >> WindowNum;
	vector<int> LastFinishTime(WindowNum,28800);	//存储每一个当前窗口结束的时间
	CustomerData TempCust;							//用以帮助入最小堆
	vector<CustomerData>Customers;
	for (int i = 0; i < CustomerNum; ++i) {
		scanf("%d:%d:%d %d", &Hour, &Minute, &Second, &Time);
		TempCust.ArriveTime = Hour * 60 * 60 + Minute * 60 + Second;
		TempCust.ProcessTime = Time * 60;
		if (TempCust.ArriveTime <= 17 * 60 * 60)			//剔除下午5点以后开始排队的顾客
			Customers.push_back(TempCust);
	}
	sort(Customers.begin(), Customers.end(), cmp);
	int CountWindowNum = 0,TotalWaitTime = 0,MinTime;
	for (unsigned int i = 0; i < Customers.size(); ++i) {
		MinTime = 100000;		//若此处MinTime初始化为86400，即24点，则最后测试点过不去，改成更大的就可以过去了
		CountWindowNum = 0;
		for (int j = 0; j < WindowNum; ++j) {
			if (MinTime > LastFinishTime[j]) {
				MinTime = LastFinishTime[j];
				CountWindowNum = j;
			}
		}
		if (Customers[i].ArriveTime > LastFinishTime[CountWindowNum]) {
			Customers[i].WaitingTime = 0;
			LastFinishTime[CountWindowNum] = Customers[i].ArriveTime + Customers[i].ProcessTime;
		}
		else {
			Customers[i].WaitingTime = LastFinishTime[CountWindowNum] - Customers[i].ArriveTime;
			LastFinishTime[CountWindowNum] += Customers[i].ProcessTime;
		}
		/*若在17点前到达的顾客等到17点还没有没服务就直接离开，那么可以加上以下两行*/
	//	if (LastFinishTime[CountWindowNum] > 17 * 60 * 60)
	//		LastFinishTime[CountWindowNum] = 17 * 60 * 60;
		TotalWaitTime += Customers[i].WaitingTime;		
	}
	if (!Customers.size()) printf("0.0\n");
	else printf("%.1lf", 1.0 * TotalWaitTime / 60 / Customers.size());
	fclose(stdin);
	system("pause");
	return 0;
}