Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1953 Accepted Submission(s): 529

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

15 2 3Sorey 3Rose 3Maltran  3Lailah 5Mikleo  61 14 21 2 3

Sample Output

Sorey Lailah Rose

Source

/*这题做了好久...题意没理解好  首先优先队列里的比较函数的大小写符号要写反..然后每个查询要先存入，再操作..并不是按顺序来的...（缺乏做题经验）  如果按之前的做法  应该是总体输入存入后 排下序搞 发现有人题解用enter[i]数组存第i个人来了可以进几个人  发现更好 所以改了下 然后操作就是  到第i个人时，都进优先队列，然后进 Min(队列里的人数，可以进的人数)   就好了... */#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
           using namespace std;const int N=150000+10;struct Node{ int v,id;}node[N],tmp;char name[N+1][200+5];int enter[N]; //第i人时可以进的人数 struct cmp{ bool operator()(Node a,Node b){ if(a.v==b.v) return a.id>b.id; return a.v
          
           ,cmp> q;int main(){ int t,k,m,p,i,now,j,a,b,z; scanf("%d",&t); while(t--){ scanf("%d%d%d",&k,&m,&p); for(i=0;i

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