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.
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.
考点明确，就是优先队列，再模拟一下可以出结果。每入队一个人就操作一次。数据量略大，用c的输入输出过了，cin，cout会TLE。
#include<stdio.h>
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue> using namespace std; struct fri
{
char name[];
int come,v;
bool operator < (const fri& f) const
{
if(v!=f.v) return v<f.v;
else return come>f.come;
} //为了优先队列里的顺序而重载
}f[]; int c[],ans[]; int main()
{
int k,m,q,t,p,sum;
int T; cin>>T; while(T--)
{
memset(c,,sizeof(c));
sum=;
cin>>k>>m>>q;
for(int i=;i<=k;++i)
{
scanf("%s %d",f[i].name,&f[i].v);
f[i].come=i;
}
for(int i=;i<=m;++i)
{
scanf("%d %d",&t,&p);
c[t]=p;
} //输入顺序不确定，t个人时进行操作 priority_queue<fri> que;
for(int i=;i<=k;++i) //每来一个人，检查是否开门
{
que.push(f[i]); //来第i个人入队
int n = c[i]; //来了i个人时应出队的n个人进行操作
while(n&&!que.empty())
{
ans[sum++]=que.top().come; //记录出队人的序号，下面用于输出名字
que.pop();
--n;
}
}
while(!que.empty()) //剩下的人依次出队
{
ans[sum++]=que.top().come;
que.pop();
} while(q--)
{
int a;
scanf("%d",&a);
printf("%s",f[ans[a]].name);
if(q==) printf("\n");
else printf(" ");
}
}
return ;
}