Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2650    Accepted Submission(s): 722

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k个人来，门先会开m次，每次开门的条件是假设当第a个人来时，可以放b个人进来。最后会把人全部放进来。查询第i个进来的是谁？

收获：题目没有说开门条件第a个人是按从小到大排序的，而且也没说a一定大于b；所以要考虑这种情况，保持程序的健壮性。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstdlib>
#include<string>
#include<map>
#include<stack>

using namespace std;
#define ll long long
#define lson le,mid,num<<1
#define rson mid+1,ri,num<<1|1
#define stop1 puts("QAQ")
#define stop2 system("pause")
#define maxn 160000

struct node
{
    char s[203];
    int val;
    int id;
    bool operator < (const node & a) const
    {
        if(val!=a.val)
            return val < a.val;
        return id>a.id;
    }

}peo[maxn];
struct kk
{
    int x,y;
}tt[maxn];

int cmp(kk a, kk b)
{
    return a.x < b.x;
}
char ans[maxn][203];

int main()
{

    int T;
    scanf("%d",&T);
    while(T--)
    {

        int k,m,q;
        scanf("%d%d%d",&k,&m,&q);
        for(int i=1;i<=k;i++)
        {
            scanf("%s%d",peo[i].s,&peo[i].val);
            peo[i].id=i;
        }

       for(int i=1;i<=m;++i)
        scanf("%d%d",&tt[i].x,&tt[i].y);

        sort(tt+1, tt+m+1,cmp);
        priority_queue<node> Q;
        memset(ans, 0, sizeof ans);

        int sum=0,pos=1;
        for(int i=1;i<=m;i++)
        {

            for(int j = pos; j <=tt[i].x; j++)
            {
                Q.push(peo[j]);
                pos++;
            }
            for(int j=1;j<=tt[i].y;j++)
            {
                if(!Q.empty())
                {node temp = Q.top();
                Q.pop();
                sum++;
                strcpy(ans[sum],temp.s);
                }
                else
                    break;
            }
        }

        for(int i=pos;i<=k;i++)
        {
            Q.push(peo[i]);
        }
        while(!Q.empty())
        {
            node temp = Q.top();
            Q.pop();
            sum++;
            strcpy(ans[sum],temp.s);
        }

        int ss;
        for(int i=1;i<q;i++)
        {
            scanf("%d",&ss);
            printf("%s ",ans[ss]);
        }
        if(q!=0)
        {
            scanf("%d",&ss);
            printf("%s\n",ans[ss]);
        }

    }
    return 0;
}