Alisha’s Party

Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1815    Accepted Submission(s): 487

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：Princess Alisha要举办party，有k个朋友参加，每个朋友到达的时间，带礼物的价值都不一样，按到达先后顺序输入朋友姓名及其所带礼物的价值，m次输入两个数t，p。代表当第t个人到达时候Alisha开门让p个人进来，当然谁带的礼物价值高谁先进，礼物价值一样时，谁先到谁先进，当然不保证m次输入有序。m次开门后，会再开一次门，让所有没进到大厅的人，依次进入。q次查询，第nq个进入Alisha大厅的是谁。输出人名之间有空格。

ps：优先队列是个好东西

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>

using namespace std;

#define maxn 150008
char name[maxn][220];

struct node
{
    int v, id;
    bool operator < (const node &t)const
    {
        if(v == t.v)
            return id > t.id;
        return v < t.v;
    }
}p[maxn];

struct people
{
    int t, p;
    bool operator < (const people &k)const
    {
        return t < k.t;
    }
}a[maxn];

int cnt[maxn];

int main()
{
    int c, k, m, q, s, g, z;
    scanf("%d", &c);

    while(c--)
    {
        g = 1;
        s = 0;
        priority_queue<node> Q;
        scanf("%d%d%d", &k, &m, &q);
        for(int i = 0; i < k; i++)   // 关系全让 下标 i 联系~
        {
            scanf("%s", name[i]);
            scanf("%d", &p[i].v);
            p[i].id = i;
        }
        for(int i = 0; i < m; i++)
            scanf("%d %d", &a[i].t, &a[i].p);
        sort(a, a+m);  // 坑点，输入的当第几个人到的开门不按顺序，要排序^^……^
        for(int i = 0; i < m; i++)
        {
            for(; s < a[i].t; s++)
                Q.push(p[s]);
            while(Q.size() && a[i].p)
            {
                node u = Q.top();
                Q.pop();
                cnt[g++] = u.id;
                a[i].p--;
            }
        }
        for(; s < k; s++)
            Q.push(p[s]);   // m次开门后如果还有人没进，一次依次进去
        while(Q.size())
        {
            node u = Q.top();
            Q.pop();
            cnt[g++] = u.id;
        }
        while(q--)
        {
            scanf("%d", &z);

            if(q >= 1)
                printf("%s ", name[cnt[z]]);
            else
                printf("%s
", name[cnt[z]]);
        }
    }
    return 0;
}