15年长春网赛Alisha’s Party +优先队列

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

n个人带着价值不同的礼物来。。
m次开门操作，，第di个人来是让p个人进去。。
（规则：价值大的先进，价值相同则先来的先进）
最后j次询问第几个进去的人是谁。。

注意两个点：
1，开门操作并不是按先后顺序，，所以要对，开门记录排序。。
2，m次开门操作后，还有人没进去，那么也要按规则进去。。

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;struct Rac{    char a[208];    int value;    int id;    bool operator < (const Rac &p) const{        if (value !=p.value) return value<p.value;        else return id>p.id;    }}people[150005];struct door{    int di,num;}m[150005];bool cmp(struct door a,struct door b){    return a.di<b.di;}priority_queue<Rac>q;struct cu{    char a[208];}out[150005];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,fuck,men;        while (!q.empty()) //清空队列        {            q.pop();        }        scanf("%d %d %d",&n,&men,&fuck);//n个人，men个开门，fuck次询问        for (int i=0;i<=n;i++)        {            memset(out[i].a,0,sizeof(out[i].a));        }        for(int i=1;i<=n;i++)        {            scanf("%s %d",people[i].a,&people[i].value);            people[i].id=i;        }        int chu=1;        int jin=1;        for(int i=1;i<=men;i++)        {            scanf("%d %d",&m[i].di,&m[i].num);        }        sort(m+1,m+men+1,cmp);        for(int i=1;i<=men;i++)        {            for(int j=chu;j<=m[i].di;j++) q.push(people[j]); //开门时第几个            for(int j=1;j<=m[i].num;j++)  //进去几人            {                if (!q.empty())                {                    Rac t=q.top();                    q.pop();                    strcpy(out[jin++].a,t.a);                }            }            chu=m[i].di+1;        }        for(int i=chu;i<=n;i++) q.push(people[i]); //所有开门操作后，没进去的也要按规则进去        while(!q.empty())        {            Rac t=q.top();            q.pop();            strcpy(out[jin++].a,t.a);        }        for(int i=1;i<=fuck;i++)        {            int aa;            scanf("%d",&aa);            if (i!=fuck) printf("%s ",out[aa].a);            else printf("%s",out[aa].a);        }        puts("");    }    return 0;}