1017 Queueing at Bank

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本文介绍了一种计算银行排队系统中客户平均等待时间的方法。考虑到银行窗口数量和服务时间限制，通过将时间转换为秒并跟踪每个客户的等待时间来求解。关键在于确定客户是否需要等待以及等待的时间长度。

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 (≤104) - 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

Analyze:

In my opinion, the key to solve these problems about time is to convert time into seconds(or minutes) from 00:00:00. To get the average waiting time of all customers, we need to know the total waiting time. The question is how to get it. It's easier to traverse every customer and record his/her waiting time. For each customer, when he arrives, there is a window available, he has no need to wait. However, when all windows are busy, he has to wait the window which is to be available earliest. The time is window[earliest] - customer.arrive_time. The solution is just like what we do in real life, but it's kind of difficult to figure it out when we deal with the problem!

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

using namespace std;

struct Customer{
	int arrive;
	int time;
};

bool Cmp(Customer a, Customer b){
	return a.arrive < b.arrive;
}

int main(){
	int n, k;
	double total = 0.0;
	int hour, minute, second, temp_min;
	vector<Customer> customer;
	cin >> n >> k;
	for(int i = 0; i < n; i++){
		Customer temp;
		scanf("%d:%d:%d %d", &hour, &minute, &second, &temp_min);
		temp.arrive = hour * 60 * 60 + minute * 60 + second;
		temp.time = temp_min * 60;
		if(temp.arrive < 61200) customer.push_back(temp);
	}
	sort(customer.begin(), customer.end(), Cmp);
	vector<int> window(k, 28800);
	for(int i = 0; i < customer.size(); i++){
		int index = 0, earliest = window[0];
		for(int j = 1; j < k; j++){
			if(window[j] < earliest){
				index = j;
				earliest = window[j];
			}
		}
		if(window[index] <= customer[i].arrive){
			window[index] = customer[i].arrive + customer[i].time;
		}else{
			total += window[index] - customer[i].arrive;
			window[index] = window[index] + customer[i].time;
		}
	}
	if(customer.size() == 0) printf("0.0");
	else printf("%.1f", total / 60 / customer.size());
	return 0;
}