1014 Waiting in Line

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:

• The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
• Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
• Customer​i​​ will take T​i​​ minutes to have his/her transaction processed.
• The first N customers are assumed to be served at 8:00am.

Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.

For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer​1​​ is served at window​1​​ while customer​2​​ is served at window​2​​. Customer​3​​ will wait in front of window​1​​ and customer​4​​ will wait in front of window​2​​. Customer​5​​ will wait behind the yellow line.

At 08:01, customer​1​​ is done and customer​5​​ enters the line in front of window​1​​ since that line seems shorter now. Customer​2​​ will leave at 08:02, customer​4​​ at 08:06, customer​3​​ at 08:07, and finally customer​5​​ at 08:10.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (≤20, number of windows), M (≤10, the maximum capacity of each line inside the yellow line), K (≤1000, number of customers), and Q (≤1000, number of customer queries).

The next line contains K positive integers, which are the processing time of the K customers.

The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.

Output Specification:

For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output Sorry instead.

Sample Input:

2 2 7 5
1 2 6 4 3 534 2
3 4 5 6 7

Sample Output:

08:07
08:06
08:10
17:00
Sorry

Solution Code:

• 如果顧客17:00之前沒有被服務，就輸出Sorry。

/***************************
每一個窗口便是一個隊列。
哪一個隊列的元素最少，黃線外的消費者優先進入哪一個隊列。
隊列未滿時允許消費者進入隊列。
******************************/
#include <iostream>
#include <queue>
#include <cstdio>
using namespace std;

int customers_minutes[1002]; //消費者的花費時間 1~1001
int windows_minutes[22] = {0}; //每個窗口的分鐘累計 1~21
int done_minutes[1002] = {0}; //消費者完成時間 1~1001
queue<int> windows[22]; // 窗口1~21

void getDoneTime(int windows_numbers, int max_numbers, int customers_numbers)
{
	// 安排最大數量的消費者進入黃線內
	int customers_id = 0; 
	for (int i = 0; i < max_numbers; ++i) {
		for (int j = 1; j <= windows_numbers; ++j) {
			windows[j].push(++customers_id);
		}
	}

	// 移除消費者，直至所有消費者
	for (int j = 0; j < customers_numbers; ++j) {
		int early_time = 99999999;
		int early_windows;
		// 計算最先辦完業務的人
		for (int i = 1; i <= windows_numbers; ++i) {
			if (!windows[i].empty() && done_minutes[windows[i].front()] == 0){
				// 當前窗口第一個消費者完成業務的時間
				done_minutes[windows[i].front()] = windows_minutes[i] + customers_minutes[windows[i].front()];
				windows_minutes[i] = done_minutes[windows[i].front()];
			}
			if (!windows[i].empty() && done_minutes[windows[i].front()] < early_time) {
				early_time = done_minutes[windows[i].front()];
				early_windows = i;
			}
		}
		
		windows[early_windows].pop(); // 移除最先辦理完業務的消費者	

		// 黃線外有人
		if (customers_id < customers_numbers) { 
			windows[early_windows].push(++customers_id);
		}
	}
}

int main()
{
	int n, m, k, q;
	cin >> n >> m >> k >> q;
	for (int i = 1; i <= k; ++i)
		cin >> customers_minutes[i];

	getDoneTime(n, m, k);
	
	// query
	int number;
	for (int i = 0; i < q; ++i) {
		cin >> number;
		// 開始時間>=540,輸出Sorry
		if (done_minutes[number] - customers_minutes[number] >= 540)
			cout << "Sorry" << endl;
		else {
			int hour = done_minutes[number] / 60;
			int minutes = done_minutes[number] % 60;
			printf("%02d:%02d\n", hour + 8, minutes);
		}
	}

	return 0;
}