PAT (Advanced Level) 1014. Waiting in Line (30) 银行排队

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

用N个队列模拟操作，每个顾客结点记录其服务时间，及自身序号。窗口当前时间加上队首顾客服务时间最小的队列pop队首顾客，push下一个顾客，同时更新当前窗口时间，根据顾客序号将出队顾客时间写入结果数组相应位置。

注意不是离队时间在17:00后才输出sorry，而是在17:00后接受服务的才输出sorry，这是个陷阱。

/*2015.7.20cyq*/#include <iostream>#include <vector>#include <string>#include <queue>#include <fstream>#include <sstream>using namespace std;//ifstream fin("case1.txt");//#define cin finstring intToString(int n,int t){if(n-t>=540)//结束服务的时间减服务时间，即开始服务的时间return "Sorry";stringstream ss;string s;int h=n/60+8;int m=n%60;if(h<10)ss<<"0";//补零ss<<h<<":";if(m<10)ss<<"0";ss<<m;ss>>s;return s;}const int MAX=2147483647;int main(){int N,M,K,Q;cin>>N>>M>>K>>Q;vector<queue<pair<int,int> > > q(N);//N个窗口，记录顾客时间和序号vector<int> custime; int x;for(int i=0;i<K;i++){//K个顾客的服务时间，这里从0开始编号cin>>x;custime.push_back(x);}int cusNum;for(cusNum=0;cusNum<M*N&&cusNum<K;cusNum++)//队列长度为M，窗口为N，最初的顾客入队q[cusNum%N].push(make_pair(custime[cusNum],cusNum));vector<int> qtime(N,0);//每个窗口的当前时间，初始化为0vector<int> result(K);//K个顾客结束服务的时间int count=0;int quickWindow;int quickTime;while(count<K){quickTime=MAX;for(int i=0;i<N;i++){//检测哪个窗口最快if(!q[i].empty()){if(qtime[i]+q[i].front().first<quickTime){quickTime=qtime[i]+q[i].front().first;quickWindow=i;}}}int tmp=q[quickWindow].front().first+qtime[quickWindow];result[q[quickWindow].front().second]=tmp;//更新结果qtime[quickWindow]=tmp;//更新最快窗口的当前时间q[quickWindow].pop();if(cusNum<K)//新顾客排队q[quickWindow].push(make_pair(custime[cusNum],cusNum));cusNum++;count++;}while(Q--){cin>>x;cout<<intToString(result[x-1],custime[x-1])<<endl;}return 0;}