hdu 5437Alisha’s Party（优先队列）

Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4384    Accepted Submission(s): 1120

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：有n个人拜访，m个操作，q个询问

每个人有个名字和带的礼物的价值，当门外有很多人时，让礼物价值大的排在前面，礼物价值相同时，让先来的人排在前面，没有两个人会同时来，题目给出来的顺序

m个操作，每个操作有个t和p代表是当第t个人来时让p个人进屋子。当最后一个人也来了的时候，所有人都可以直接进屋子。

q个询问，每个询问给出一个编号i，求第i个进屋子的人的名字

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 150010
struct Node
{
    char name[210];
    int id;
    long long val;
} e[N];
bool operator <(Node a,Node b)
{
    if(a.val==b.val) return a.id>b.id;
    return a.valq;
int ans[N];
int main()
{
    int T,t,p;
    int n,m,Q,now,cnt;
    scanf("%d",&T);
    while(T--)
    {
        now=1;
        cnt=1;
        scanf("%d %d %d",&n,&m,&Q);
        for(int i=1; i<=n; i++)
        {
            scanf("%s %lld",e[i].name,&e[i].val);
            e[i].id=i;
        }
        for(int i=1; i<=m; i++)
        {
            scanf("%d %d",&t,&p);
            for(int i=now; i<=t; i++)
            {
                if(now>n) continue;
                q.push(e[now++]);
            }
            if(now>n)
                while(!q.empty())
                {
                    Node w=q.top();
                    ans[cnt++]=w.id;
                    q.pop();
                }
            else
            {
                for(int i=1; i<=p; i++)
                {
                    if(!q.empty())
                    {
                        Node w=q.top();
                    ans[cnt++]=w.id;
                    q.pop();
                    }
                }
            }
        }
        while(!q.empty())
                {
                    Node w=q.top();
                    ans[cnt++]=w.id;
                    q.pop();
                }
        for(int i=1;i<=Q;i++)
        {
            scanf("%d",&t);
            printf("%s",e[ans[t]].name);
            if(i!=Q) printf(" ");
            else printf("\n");
        }
    }
    return 0;
}