1014. Waiting in Line (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：

1.输入N（窗口数），M（每个窗口黄线内最大容量），K（总共的顾客数，编号1-K），Q（要查询自己服务结束时间的Q个顾客）

2.输入K个数，每个顾客服务所需时间，输入Q个顾客号

3.窗口内人数小于最大容量时，黄线外顾客才能选择队伍最短的进入等待

4.Q个顾客号，依次按行输出其服务结束的时间，若在17：00前顾客不能开始其服务，则输出Sorry

思路：

利用有限队列，以服务结束时间早为高优先级，一开始，先将N*M个顾客分配进各自的窗口等待，之后，每次pop一个顾客，就可以在该窗口加入一个顾客。直至所有顾客进入黄线内等待。

注意：

输出的时间格式，HH：MM, 小时和分钟都是两位数；输出Sorry是顾客开始服务的时间大于17：00

#include<iostream>#include<cstring>#include<algorithm>#include<queue>using namespace std;struct node{  // int cusNum;  int win;//窗口号  int stime;//该顾客结束服务的时间  friend bool operator < (node a,node b)//结束服务时间小的优先级高  {    return a.stime>b.stime;  }}custom[1005],top;int main(){  int N,M,K,Q;  int windows[30];//窗口对应的已花费时间  int ptime[1005];//顾客所需服务时间  int queries[1005];//要求查询的顾客号  int c=0;//顾客号计数  //int stime[1005];  priority_queue<node>q;  cin>>N>>M>>K>>Q;  for(int i=1;i<=K;i++)    cin>>ptime[i];  for(int i=0;i<Q;i++)    cin>>queries[i];  for(int i=1;i<=N;i++)    windows[i]=0;  for(int i=0;i<M;i++)    {      for(int j=1;j<=N;j++){  if(c>K)break;  c++;  // custom[c].cusNum=c;  custom[c].win=j;  windows[j]+=ptime[c];  custom[c].stime=windows[j];//顾客结束服务时间是窗口以花费时间加上其所需的服务时间  q.push(custom[c]);}    }  // cout<<"c:"<<c<<endl;  while(c<K)    {      top=q.top();      q.pop();      //cout<<top.cusNum<<" "<<top.win<<" "<<top.stime<<"\t";      c++;      //custom[c].cusNum=c;      custom[c].win=top.win;//一个顾客走了，才能有新顾客进入      windows[top.win]+=ptime[c];      custom[c].stime=windows[top.win];      q.push(custom[c]);    }  // cout<<endl;  for(int i=0;i<Q;i++)    {      if(custom[queries[i]].stime-ptime[queries[i]]>=540)//顾客开始服务时间=顾客结束服务时间-其所需服务时间{  cout<<"Sorry\n";  continue;}      int HH=custom[queries[i]].stime/60+8;      int MM=custom[queries[i]].stime%60;      if(HH<10)cout<<0;      cout<<HH<<":";      if(MM<10)cout<<0;      cout<<MM<<endl;    }  return 0;}