1017 Queueing at Bank

最新推荐文章于 2023-12-27 11:45:21 发布

原创最新推荐文章于 2023-12-27 11:45:21 发布 · 231 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲级专栏收录该内容

35 篇文章

订阅专栏

本文介绍了一个银行排队系统的模拟案例，通过算法计算在特定营业时间内，多个窗口服务下顾客的平均等待时间。考虑到银行营业时间限制、顾客到达时间及服务时间上限，文章详细解析了算法实现步骤。

1017 Queueing at Bank （25 分）

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

题型分类：快乐模拟

题目大意：银行的营业时间为8:00 - 17:00，给定K个窗口，顾客依次在线后等待，计算被服务到的（17:00以后的顾客没有被服务到）顾客的平均等待时间。

解题思路：注意几个细节就好：①每个顾客的服务时间不超过1小时。②银行8:00以后才开始营业。③17:00以后的顾客不会被服务到，不计入数据。

#include <cstdio>
#include <algorithm>

using namespace std;

const int maxCus = 10010;
const int maxWin = 110;
const int INF = 0x3f3f3f3f;

typedef struct {
	int arriveTime;
	int processTime;
}customer;

customer cus[maxCus];
int window[maxWin];

int convertToSecond(int hh, int mm, int ss);
bool cmp(customer a, customer b);

int main(int argc, char** argv) {
	int N, K;
	int stTime = convertToSecond(8, 0, 0), edTime = convertToSecond(17, 0, 0);
	scanf("%d %d", &N, &K);
	for(int i = 0; i < N; i++){
		int hh, mm, ss;
		scanf("%d:%d:%d %d", &hh, &mm, &ss, &cus[i].processTime);
		cus[i].processTime = cus[i].processTime > 60 ? 60 * 60 : cus[i].processTime * 60; //一个客户最多服务1H 
		cus[i].arriveTime = convertToSecond(hh, mm, ss);
	}
	sort(cus, cus + N, cmp);
	int totalWait = 0;
	for(int i = 0; i < K; i++){
		window[i] = stTime;
	}
	int cnt = 0;
	for(int i = 0; i < N; i++){
		if(cus[i].arriveTime > edTime) break; 
		int winNum = -1, min = INF;
		for(int j = 0; j < K; j++){
			if(window[j] < min){
				winNum = j;
				min = window[j];
			}
		}
		if(window[winNum] < cus[i].arriveTime){ //此时表示顾客不用等待 
			window[winNum] = cus[i].arriveTime + cus[i].processTime;
		}else{
			totalWait += window[winNum] - cus[i].arriveTime;
			window[winNum] = window[winNum] + cus[i].processTime;
		}
		cnt++;
	}	
	printf("%.1f", (double)totalWait / 60 / cnt);
	
	return 0;
}

int convertToSecond(int hh, int mm, int ss){
	return hh * 3600 + mm * 60 + ss; 
}

bool cmp(customer a, customer b){
	return a.arriveTime < b.arriveTime;
}