PAT1014

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


模拟银行排队的题，不是很难，我的思想是：先让客户填满窗口，然后每一分钟都进行一次判断，看是否有人办理完业务，是否有人需要进入待办理业务区域，
一直到17:00，然后处理那些已经开始但是还未完成的客户（注意这点，我第一次有几个case没过，就是因为这个），最后结果存在一个键值对map里，查询即可。

#include<iostream>
#include<queue>
#include<list>
#include<vector>
#include<map>
#include<iomanip>
using namespace std;

struct Node
{
    int id;
    int ptime;
    int etime;
    int served;
};

int main()
{
    int n,M,k,q;
    while(cin>>n>>M>>k>>q)
    {
        vector<list<Node>> windows(n);
        queue<Node> wait;
        vector<int> check(q);
        map<int, Node> done;
        for(int i=0; i<k; ++i)
        {
            int a; cin>>a;
            Node n={i+1, a, 0, 0};
            wait.push(n);
        }
        for(int i=0; i<q; ++i)
            cin>>check[i];
        int time = 0;

        for(int i=0; i<M; ++i)
            for(int j=0; j<n; ++j)
                if(windows[j].size() < M && wait.size() != 0)
                {
                    windows[j].push_back(wait.front());
                    wait.pop();
                }
        /*窗口预处理*/
        while(time < 540 )
        {
            ++time;
            for(int i=0; i<windows.size(); ++i)
                if(windows[i].size() != 0)
                {
                    --windows[i].begin()->ptime;
                    windows[i].begin()->served = 1;
                    if( windows[i].begin()->ptime== 0)
                    {
                        windows[i].begin()->etime = time;
                        done[windows[i].begin()->id] = *(windows[i].begin());
                        windows[i].pop_front();
                    }
                }
            /*处理窗口前排客户*/

            for(int i=0; i<windows.size(); ++i)
                if(wait.size() != 0 && windows[i].size() < M )
                {
                    windows[i].push_back(wait.front());
                    wait.pop();
                }
            /*若有空位置，处理排队客户*/
        }
        for(int i=0; i<windows.size(); ++i)
            if(windows[i].size() != 0)
                if(windows[i].begin()->served == 1)
                {
                    windows[i].begin()->etime = time + windows[i].begin()->ptime;
                    done[windows[i].begin()->id] = *(windows[i].begin());
                }
        /*下班时间到，处理前台已经开始但是未完成的客户*/

        for(int i=0; i<check.size(); ++i)
            if(done.find(check[i]) != done.end())
            {
                int hour = done[check[i]].etime/60;
                int minutes = done[check[i]].etime%60;
                cout<<setfill('0')<<setw(2)<<hour+8<<":"<<setw(2)<<minutes<<endl;
            }
            else
                cout<<"Sorry"<<endl;
    }
    return 0;
}