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

需要考虑的有队列头的等待时间在8:00-17:00之间但处理时间加上等待时间超出17:00也是可以处理的，测试点2 4 5与这个有关

#include <iostream>#include <queue>#include <cstdio>#define MAX 1005using namespace std;typedef struct customer{    int costtime;    int waittime;}cust;cust c[MAX] = {0};const int maxtime = 9*60;const int starttime = 8*60;int main(){    int N, M, K, Q;    int counter = 0, numinque = 0;    int windows[MAX] = {0};    cin >> N >> M >> K >> Q;    queue<cust*> v1[N];    for(int i=0; i<K; i++)    {        cin >> c[i].costtime;        c[i].waittime = -1;        if(v1[N-1].size() < M)        {            numinque++;            v1[counter++].push(&c[i]);//元素依次入队            if(counter == N)                counter = 0;        }    }    for(int j=0; j<N; j++)    {        if(v1[j].empty())                continue;        (v1[j].front())->waittime = 0;//开始的队里第一个元素一定能处理等待时间为0处理时间为其costtime    }    for(int i=1; i<=maxtime; i++)    {        for(int j=0; j<N; j++)        {            windows[j]++;//记录每个队列的第一位处理时间            if(v1[j].empty())                continue;            if(((v1[j].front())->costtime-windows[j]) == 0)//如果处理时间和他所需花费时间相等即处理完毕则出队            {                windows[j] = 0;                (v1[j].front())->waittime = i;//处理完毕的等待时间                (v1[j].front())->costtime = 0;                v1[j].pop();                v1[j].push(&c[numinque++]);//下一位入队                if(i != maxtime)//如果未到最后时刻的人都可以处理                    (v1[j].front())->waittime = i;//等待时间置为i;            }        }    }    int custnum, time, h, m;    while(Q--)    {        cin >> custnum;        custnum--;        if(c[custnum].waittime == -1)            cout << "Sorry" << endl;        else        {            time = 0;            time += c[custnum].waittime+c[custnum].costtime+starttime;            h = time/60;            m = time-h*60;            printf("%02d:%02d\n", h, m);        }    }    return 0;}