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 51 2 6 4 3 534 23 4 5 6 7

Sample Output

08:0708:0608:1017:00Sorry

---

这是个排队的问题，可以用队列来做。有n个窗口，也就是可以设置n个队列，每个队列最长为m人。首先让n*m个人入队（如果有这么多人）。然后对于后面的每个人，挑全部队列中第一个人剩余时间最短（最短时间设为Min）的那个队列，这个队列出队一个人，然后当前的这个人入队，同时更新第一排每个人的剩余时间（用当前剩余时间减去Min）。入队的同时更新每个队列最后一个人完成时的时间和每个人入队的时间（等于原本的最后一个人完成的时间）。最后对于查询的某个人，如果它的开始时间小于540，说明能接受服务，输出它完成的时间（开始时间加花费的时间），否则说明不能接受服务，输出“Sorry”。

代码：

#include <iostream>#include <cstring>#include <vector>#include <cstdlib>#include <cstdio>#include <queue>using namespace std;int main(){int n,m,k,q;cin>>n>>m>>k>>q;vector<queue<int> >que(n);vector<int>time(n,0);vector<int>start_time(k,0);vector<int>cost_time(k);int index=0,count=0;for(int i=0;i<k;i++){int t;cin>>t;cost_time[i]=t;if(count<m*n){start_time[i]=time[index];time[index]+=t;que[index].push(t);index=(index+1)%n;count++;}else{int Min=1<<30;for(int j=0;j<n;j++){if(que[j].front()<Min) {Min=que[j].front();index=j;}}for(int j=0;j<n;j++){que[j].front()-=Min;}start_time[i]=time[index];time[index]+=t;que[index].pop();que[index].push(t);}}for(int i=0;i<q;i++){int target;cin>>target;int t=start_time[target-1];if(t<540) {t+=cost_time[target-1];printf("%02d:%02d\n",8+t/60,t%60);}else printf("Sorry\n");}}