﻿ 1014 Waiting in Line (30)（30 分）

### 1014 Waiting in Line (30)（30 分）

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 customers 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

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
```优先队列。首先前n*m个人可以被直接容纳，紧接着就要看哪个队的第一个人先完成，那么这一队就空出一个位置，然后，就可以再插一个人，那么这一队原来的第二人就变成了本队的第一人，依次下去，每次都要知道所有队的第一人谁先办理完业务，所以用到优先队列。代码：```
```#include <bits/stdc++.h>

using namespace std;
int n,m,k,q;
int qq;
int s[21],r[21];///s记录每一队最后一人办完的时间总数，方便之后新插入人的时间总数 r记录第一人办完的时间总数，方便知道是否有空可以插人
int c[1001],need[1001];///c记录每个人办理结束的总的时间，need记录从开始办理到结束办理花费的时间，只要 开始办理的时间在17点之前就可以
typedef pair<int,int> P;
void print(int t)
{
if(c[t] - need[t] >= 540)puts("Sorry");
else printf("%02d:%02d
",8 + c[t] / 60,c[t] % 60);
}
int main()
{
scanf("%d%d%d%d",&n,&m,&k,&q);
queue<int> que[21];
priority_queue<P,vector<P>,greater<P> > pq;
for(int i = 0;i < min(k,n * m);i ++)
{
scanf("%d",&need[i]);
que[i % n].push(need[i]);
c[i] = (s[i % n] += need[i]);
if(i >= n * (m - 1))
{
pq.push(P(que[i % n].front(),i % n));
que[i % n].pop();
}
}
for(int i = 0;i + n * m < k;i ++)
{
scanf("%d",&need[i + n * m]);
int present = pq.top().second,time = pq.top().first;
pq.pop();
que[present].push(need[i + n * m]);
s[present] += need[i + n * m];
c[i + n * m] = s[present];
time += que[present].front();
que[present].pop();
pq.push(P(time,present));
}
for(int i = 0;i < q;i ++)
{
scanf("%d",&qq);
print(qq - 1);
}
}```
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