PAT-1014 Waiting in Line (30 分) 优先队列

标签：start   pos   using   operator   win   sse   struct   ons   final   

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 (*N**M*+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.
• Customeri 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. *Custome**r*5 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 *custome**r*5 at 08:10.

Input Specification:

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 Specification:

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

思路分析 先把前n*m个依此放进队列,每次优先队列弹出一个就再放一个即可

此题我犯了好多错误

1.优先队列取了一个节点居然忘记pop了,和de了老半天

2.审题错误,问的是开始时间在17:00之前的,不是服务结束后在17:00之前的,而且样例给了17:00也算,所以应该是

if(start_time[query]>=540)///审题

? {

? printf("Sorry\n");

? }

#include<bits/stdc++.h>
#define de(x) cout<<#x<<" "<<(x)<<endl
#define each(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
const int maxn=1000+5;
const int inf=0x3f3f3f3f;
struct node
{
    int time,id;
    node(int id=0,int time=0):id(id),time(time){}
    friend bool operator<(node a,node b)
    {
        if(a.time==b.time)
            return a.id>b.id;
        else return a.time>b.time;
    }
};
int wait_time[maxn];
int deadline[maxn];
int start_time[maxn];
int ans[maxn];
priority_queue<node>Q;
/*
2 2 7 7
1 2 6 4 3 534 2
1 2 3 4 5 6 7
*/
int main()
{
    int n,m,k,q;
    scanf("%d%d%d%d",&n,&m,&k,&q);
    //de(k);
    each(i,1,k)scanf("%d",&wait_time[i]);
    //de(wait_time[1]);
    int poi=1;
    each(i,1,m)
    {
        each(j,1,n)
        {
            Q.push(node(j,deadline[j]+wait_time[poi]));
            //de(poi);
            start_time[poi]=deadline[j];
            ans[poi]=deadline[j]+wait_time[poi];
            //de(deadline[j]);
            //de(wait_time[poi]);
            //de(ans[poi]);
            deadline[j]+=wait_time[poi];
            poi++;
        }
    }
    poi=m*n+1;
    node temp;
    while(poi<=k)
    {
        temp=Q.top();
        Q.pop();//居然忘记了
        int id=temp.id;
        //de(poi);
        //de(id);
        //de(deadline[id]);
        int time=temp.time;
        //de(time);
        start_time[poi]=deadline[id];
        ans[poi]=deadline[id]+wait_time[poi];
        Q.push(node(id,ans[poi]));
        deadline[id]+=wait_time[poi];
        poi++;
    }
    while(q--)
    {
        int query;
        cin>>query;
        //de(ans[query]);
        //de(start_time[query]);
        if(start_time[query]>=540)///审题
        {

            printf("Sorry\n");
        }
        else
            printf("%02d:%02d\n",8+ans[query]/60,ans[query]%60);
    }
    return 0;
}

PAT-1014 Waiting in Line (30 分) 优先队列

标签：start   pos   using   operator   win   sse   struct   ons   final   

原文地址：https://www.cnblogs.com/Tony100K/p/11758028.html