1017. Queueing at Bank (25)

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 (<=10000) - 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

/*
 * 银行排队没有涉及队列，更多的是排序
 */
#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
using namespace std;
struct Customer
{
	int come;
	int done;
	bool operator < (const Customer &cus) const
	{
		return come < cus.come;
	}
};
vector<Customer> customer;
vector<int> windows;
int main(void)
{
	int n, k;
	scanf("%d%d", &n, &k);
	customer.resize(n);
	windows.resize(k, 8 * 60 * 60);
	int hh, mm, ss, done;
	for (int i = 0; i < n; ++i)
	{
		scanf("%d:%d:%d%d", &hh, &mm, &ss, &done);
		customer[i].done = done > 60 ? 3600 : done * 60;
		customer[i].come = hh * 3600 + mm * 60 + ss;
	}
	sort(customer.begin(), customer.end());
	int serverN = 0;
	int serverW = 0;// 24 * 60 * 60 * 10000 < 10^9
	int deadTime = 17 * 60 * 60;
	for (int i = 0; i < n; ++i)
	{
		//寻找能够服务的窗口，如果没有寻找等待事件最少的窗口
		if (customer[i].come > deadTime) //题目说的是超过17:00过来的不服务，其他的正在等待的一定要服务，不过是否已经到了下班时间
			continue;
		int win = 0;
		for (int j = 0; j < k; ++j)
		{
			if (windows[j] <= customer[i].come) //寻找可以服务的的窗口
			{
				win = j;
				break;
			}
			else
			{
				if (windows[j] < windows[win]) //寻找等待事件最小的窗口
					win = j;
			}
		}
		++serverN;
		if (windows[win] <= customer[i].come)
		{
			windows[win] = customer[i].come + customer[i].done;
			
		}
		else
		{
			serverW += windows[win] - customer[i].come;
			windows[win] += customer[i].done;
		}
	}
	printf("%.1f\n", serverN > 0 ? serverW * 1.0 / (serverN * 60) : 0);
	return 0;
}