PAT Advanced 1014 Waiting in Line

Suppose a bank hasN 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 Mcustomers. Hence when all theN lines are full, all the customers after (and including) the(st one will have to wait in a line behind the yellow line.< /span>
  • 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< span class="vlist">?i?? minutes to have his/her transaction processed.< /span>
  • The firstN customers are assumed to be served at 8:00am.

Now given the processing time of each custome r, 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, customer?1??is served atwindow?1??whilecu stome r?2?? is served at winddow?2 ??. Customer?3?? will wait in front ofwindo w?1??andcustomer?4?? will wait in front ofwindow?2 ??. Customer ?5??will wait behind the yellow line. span> span> span> span> span> span>

At 08:01, custoomer?1?? is done and customer?5??enters the line in front of window?1?? Since that line seems shorter now. Custome r?2??will leave at 08:02, customer ?4< span class="fontsize-ensurer reset-size5 size5">?? at 08:06, customer?3?? at 08:07, an d finally customer?5 ?? at 08:10.< /span>< /span> span> span> span>

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers: < span class="base textstyle uncramped">N (≤, number of window s), M (≤, the maximum capacity of each line inside the yellow line), < span class="strut">K (≤, number of customers ), and Q (≤, number of customer queries).< /span>

The next line contains K positive integers, which are the processing time of the K customers. span>

The last line contains < span class="katex-html">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 < span class="strut bottom">K.< /span>

Output Specification:

For each of theQ 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:



Maintain N window queues, in which the number of each person and the time of entering the queue are stored (The number of minutes that have passed since 8:00). When everyone arrives, look for a queue that can be entered. If found, enter the queue directly. If all queues are full, calculate the completion time of the head of each queue, and remove the person who completed the earliest and join the new team. People coming. When everyone arrives, calculate the dequeue time of everyone in the remaining queue. Finally, exclude those who joined the team later than 17:00, and output the result. 08:07
08:06
08:10
17:00
Sorry
#include 

using namespace std;
struct cmp
{
bool operator () (pair<int,int> a,pair<int ,int> b)
{
if(a.second==b.second)
{
return a.first>b.first;
}
else
{
return a.second>b.second;
}

}
};
int main()
{
int N,M,K,Q,t;

cin
>>N>>M>>K>>Q;
vector
<int> p_times(K+1), f_times(K+1);
vector
int,int> >> winds(N,queueint,int> >());
queue
<int> outside;
priority_queue
int,int>,vector int,int> >,cmp> mtime;
for(int i=1;i<=N;i++)
mtime.push(make_pair(i
-1,0));
for(int i=1;i<=K;i++)
{
//cout<
cin>>t;
p_times[i]
=t;
bool hasIn=false;
int mincnt=M,minptr=-1< span style="color: #000000;">;
for(int j=0;j)
{
if(winds[j].size()<M)
{
if(mincnt>winds[j].size())
{
mincnt
=winds[j].size();
minptr
=j;
}

}
}
if(minptr!=-1)
{
winds[minptr].push(make_pair(i,
0)) ;
hasIn
=true;
}


if(!hasIn)
{
priority_queue
int,int>,vector int,int> >,cmp> tmp;
for(int j=0;j)
{
if(!winds[j].empty())
tmp.push(make_pair(j,p_times[winds[j].front().first]
+winds[j].front() .second));
}
pair
<int,int> ans=tmp.top();
//cout<
f_times[winds[ans.first].front().first]=ans.second;
winds[ans.first].pop();
winds[ans.first].push(make_pair(i,ans.second));
winds[ans.first].front().second
=ans.second;
}
}
int cnt=0;
for(int i=0;i)
{
cnt
+=winds[i].size();
}
//cout<
for(int i=0;i)
{
priority_queue
int,int>,vector int,int> >,cmp> tmp;
for(int j=0;j)
{
if(!winds[j].empty())
tmp.push(make_pair(j,p_times[winds[j].front().first]
+winds[j].front() .second));
}
pair
<int,int> ans=tmp.top();
//cout<
f_times[winds[ans.first].front().first]=ans.second;
winds[ans.first].pop();
winds[ans.first].front().second
=ans.second;
}
for(int i=1;i<=Q;i++)
{
cin
>>t;
int HH=f_times[t]/60+ 8;
int MM=f_times[t]%60< span style="color: #000000;">;
int pre=f_times[t]-p_times[t];
int PHH=pre/60+8;
int PMM=pre%60;
if(PHH>=17)
{
cout
<<"Sorry"<<endl;
}
else
{
printf(
"%02d:%02d ",HH,MM);
}

}
return 0;
}

#include 

using namespace std;
struct cmp
{
bool operator () (pair<int,int> a,pair<int ,int> b)
{
if(a.second==b.second)
{
return a.first>b.first;
}
else
{
return a.second>b.second;
}

}
};
int main()
{
int N,M,K,Q,t;

cin
>>N>>M>>K>>Q;
vector
<int> p_times(K+1), f_times(K+1);
vector
int,int> > > winds(N,queueint,int> >());
queue
<int> outside;
priority_queue
int,int>,vectorint,int> >,cmp> mtime;
for(int i=1;i<=N;i++)
mtime.push(make_pair(i
-1,0));
for(int i=1;i<=K;i++)
{
//cout<
cin>>t;
p_times[i]
=t;
bool hasIn=false;
int mincnt=M,minptr=-1;
for(int j=0;j)
{
if(winds[j].size()<M)
{
if(mincnt>winds[j].size())
{
mincnt
=winds[j].size();
minptr
=j;
}

}
}
if(minptr!=-1)
{
winds[minptr].push(make_pair(i,
0));
hasIn
=true;
}


if(!hasIn)
{
priority_queue
int,int>,vectorint,int> >,cmp> tmp;
for(int j=0;j)
{
if(!winds[j].empty())
tmp.push(make_pair(j,p_times[winds[j].front().first]
+winds[j].front().second));
}
pair
<int,int> ans=tmp.top();
//cout<
f_times[winds[ans.first].front().first]=ans.second;
winds[ans.first].pop();
winds[ans.first].push(make_pair(i,ans.second));
winds[ans.first].front().second
=ans.second;
}
}
int cnt=0;
for(int i=0;i)
{
cnt
+=winds[i].size();
}
//cout<
for(int i=0;i)
{
priority_queue
int,int>,vectorint,int> >,cmp> tmp;
for(int j=0;j)
{
if(!winds[j].empty())
tmp.push(make_pair(j,p_times[winds[j].front().first]
+winds[j].front().second));
}
pair
<int,int> ans=tmp.top();
//cout<
f_times[winds[ans.first].front().first]=ans.second;
winds[ans.first].pop();
winds[ans.first].front().second
=ans.second;
}
for(int i=1;i<=Q;i++)
{
cin
>>t;
int HH=f_times[t]/60+8;
int MM=f_times[t]%60;
int pre=f_times[t]-p_times[t];
int PHH=pre/60+8;
int PMM=pre%60;
if(PHH>=17)
{
cout
<<"Sorry"<<endl;
}
else
{
printf(
"%02d:%02d ",HH,MM);
}

}
return 0;
}

Leave a Comment

Your email address will not be published.