HDU 5437 Alisha’s Party

最新推荐文章于 2021-03-13 08:02:31 发布

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最新推荐文章于 2021-03-13 08:02:31 发布

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分类专栏： HDU 文章标签： hdu 优先队列

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阿丽莎邀请了多位朋友参加她的生日派对，但派对场地有限，她只能让部分人同时进入。通过优先队列的模拟方式，决定谁将先进入派对并放下礼物，最终回答了q次询问，告知第n个人是谁。此过程涉及到资源分配的策略和公平性考虑。

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Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2736 Accepted Submission(s): 741

Problem Description

Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value

v , and all of them will come at a different time. Because the lobby is not large enough, Alisha can only let a few people in at a time. She decides to let the person whose gift has the highest value enter first.

Each time when Alisha opens the door, she can decide to let

p people enter her castle. If there are less than

p people in the lobby, then all of them would enter. And after all of her friends has arrived, Alisha will open the door again and this time every friend who has not entered yet would enter.

If there are two friends who bring gifts of the same value, then the one who comes first should enter first. Given a query

n Please tell Alisha who the

n−th person to enter her castle is.

Input

The first line of the input gives the number of test cases,

T , where

1≤T≤15 .

In each test case, the first line contains three numbers

k,m and

q separated by blanks.

k is the number of her friends invited where

1≤k≤150,000 . The door would open m times before all Alisha’s friends arrive where

0≤m≤k . Alisha will have

q queries where

1≤q≤100 .

The

i−th of the following

k lines gives a string

Bi , which consists of no more than

200 English characters, and an integer

vi ,

1≤vi≤108 , separated by a blank.

Bi is the name of the

i−th person coming to Alisha’s party and Bi brings a gift of value

vi .

Each of the following

m lines contains two integers

t(1≤t≤k) and

p(0≤p≤k) separated by a blank. The door will open right after the

t−th person arrives, and Alisha will let

p friends enter her castle.

The last line of each test case will contain

q numbers

n1,...,nq separated by a space, which means Alisha wants to know who are the

n1−th,...,nq−th friends to enter her castle.

Note: there will be at most two test cases containing

n>10000 .

Output

For each test case, output the corresponding name of Alisha’s query, separated by a space.

Sample Input

  
   1
5 2 3
Sorey 3
Rose 3
Maltran  3
Lailah 5
Mikleo  6
1 1
4 2
1 2 3

Sample Output

  
   Sorey Lailah Rose

Source

2015 ACM/ICPC Asia Regional Changchun Online

///题意:优先队列模拟阿丽莎过生日邀请了k个朋友去做客 ,但是她卧室太小了所以先让他们到客厅每个朋友都带着价值v的礼物来的 ,阿丽莎会开卧室门m次当第t个人来了后她会带p个人进卧室把礼物放下
///规则@: 礼物价值大的先进相同价值的先来的先进
///最后一次开门后如果还有没放下礼物的按照同样规定进卧室放礼物,q次询问,询问第i个进入卧室的人是谁
///坑点1:开卧室门m次,需要对m个人进行按编号排序
///坑点2:开卧室门m次后,如果仍有人没进入卧室,则应按之前的规则@进入
///误区:如果每次开门,所允许进入的人数大于当前人数,则当前空出来的允许进入权并不算在下一次开门所允许进入人数中

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
    int num;
    int value;
    char name[210];
}peo[150010];
struct nodo
{
    int t;
    int p;
}per[150010];
int numname[150010];
int ans;
bool operator < (const node &t1,const node &t2)
{
    if(t1.value==t2.value)  
    {
        return t1.num>t2.num; /// 按编号从小到大排序,编号小的先出队列
    }
    else return t1.value<t2.value; /// 按价值从大到小排序,m价值大的先出队列
}

bool cmp(nodo x,nodo y)
{
    return x.t<y.t;  
}

int main()
{
    int T;
    int k,m,q,n;
    int i,j;
    scanf("%d",&T);
    while(T--)
    {
        priority_queue<node>qu;  ///优先队列
        ans=0;
        scanf("%d%d%d",&k,&m,&q);
        for(i=1;i<=k;i++)
        {
            scanf("%s%d",peo[i].name,&peo[i].value);
            peo[i].num=i;
        }
        memset(per,0,sizeof(per));
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&per[i].t,&per[i].p);
        }
        sort(per,per+m+1,cmp);
        for(i=1;i<m;i++)
        {
            if(per[i].t==per[i+1].t)
            {
                per[i].t+=per[i+1].t;
                for(j=i+1;j<m;j++)
                {
                    per[j]=per[j+1];
                }
                m--;
                i--;
            }
        }
        memset(numname,0,sizeof(numname));
        for(i=1;i<=m;i++)
        {
            for(j=per[i-1].t+1;j<=per[i].t;j++)
            {
                qu.push(peo[j]);
            }
            while(!qu.empty())
            {
                node x=qu.top();
                numname[++ans]=x.num;
                per[i].p--;
                qu.pop();
                if(per[i].p==0)  break;
            }
        }
        for(j=per[m].t+1;j<=k;j++)
        {
            qu.push(peo[j]);
        }
        while(!qu.empty())
        {
            node x=qu.top();
            numname[++ans]=x.num;
            qu.pop();
        }
        int flag=0;
        while(q--)
        {
            scanf("%d",&n);
            if(flag==0)
            {
                flag=1;
               printf("%s",peo[numname[n]].name);
            }
            else printf(" %s",peo[numname[n]].name);
        }
        printf("\n");
    }
    return 0;
}