PAT 1014 Waiting in Line (30)



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, customer₁ is served at window₁ while customer₂ is served at window₂. Customer₃ will wait in front of window₁ and customer₄ will wait in front of window₂. Customer₅ will wait behind the yellow line.

At 08:01, customer₁ is done and customer₅ enters the line in front of window₁ since that line seems shorter now. Customer₂ will leave at 08:02, customer₄ at 08:06, customer₃ at 08:07, and finally customer₅ at 08:10.

Input

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

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

/* 题目大意：程序模拟银行排队办业务。
   共n个窗口 每个窗口外都有黄线 黄线内最多可以站m个人进行排队 共有k个人 剩下的k-m*n个人就只能在黄线外等待。每当黄线内可以站人时，黄线外的人就进入需要排队的人最少的窗口进行排队（若存在多个这样的窗口，则选择窗口序号最小的）。17:00之前没能开始办理业务的 输出Sorry 否则输出办完业务的时间。 
*/
#include "iostream"
#include "stdio.h"
#include "cstring"
#include "vector"
#include "queue"
using namespace std;
#define INF 1000
int main() {
	int timeBase[1001];  /* 办理业务所需要的时间 */
	int preLeaveTime[21]; /* 轮到该下一个人办业务时的时间 */
	int leaveTime[1001]; /* 办理完业务时的时间 */
	int n, m, k, q;

	cin >> n >> m >> k >> q;
	for (int i = 0; i < k; i++) {
		cin >> timeBase[i];
	}
	memset(preLeaveTime, 0, sizeof(preLeaveTime));
	vector<queue<int>>window(n+1); /* 用队列数组模拟n个窗口 */
	int i;
	for ( i = 0; i < m*n && i<k; i++) { /* 确定最开始的m*n || k个人的进入黄线的时间 和进入了哪个窗口 */
		leaveTime[i] = preLeaveTime[i%n] + timeBase[i];
		preLeaveTime[i%n] = leaveTime[i]; 
		window[i%n].push(i); /* 将第i个人 放入第i%n条队列 */
	}
	for (; i < k; i++) {
		int Min = INF;
		int MinX = -1;
		for (int j = 0; j < n; j++) { /* 按序号从小到大遍历所有的队列 找出preLeaveTime最小的 将其pop出队列 从黄线外选择一个进队 */
			int top = window[j].front();
			if (leaveTime[top] < Min) {
				Min = leaveTime[top];
				MinX = j;
			}
		}
			window[MinX].pop();
			window[MinX].push(i);
			leaveTime[i] = preLeaveTime[MinX] + timeBase[i];
			preLeaveTime[MinX] = leaveTime[i];
	}
	while (q--) {
		int query;
		scanf("%d", &query);
		if (leaveTime[query-1] - timeBase[query-1] >= 540)
			printf("Sorry\n");
		else
		printf("%02d:%02d\n", 8+leaveTime[query-1]/60, leaveTime[query-1]%60);
	}
	return 0;
}