HDU5437(2015长春网赛)——Alisha’s Party（优先队列）

Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5309    Accepted Submission(s): 1337

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. Biis 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

 

裸地优先队列。每次开门的时候把人压到优先队列里，然后搞出value最高得几个人加到结果数组里头。

注意千万不能用cin和cout，被坑惨了。会output limit exceed

#include <cmath>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <ctime>
#include <cstdlib>
#include <iostream>
using namespace std;
#define MAXN 200010
#define LEN 200010
#define INF 1e9+7
#define MODE 1000000
#define pi acos(-1)
#define g 9.8
typedef long long ll;

struct node{
    char name[205];
    int v;
    int num;
};
bool operator < (const node &a,const node &b){
    if(a.v==b.v)
        return a.num>b.num;
    return a.v<b.v;
}
node a[MAXN];
char res[MAXN][205];
int k,m,q;
priority_queue <node> que;
struct op{
    int t,p;
};
op o[MAXN];
bool cmp(op a,op b){
    return a.t<b.t;
}
int main()
{
    //freopen("in.txt","w",stdin);
    //std::ios::sync_with_stdio(false);
    int T;
    scanf("%d",&T);
    while(T--){
        while(!que.empty()){
            que.pop();
        }
        scanf("%d%d%d",&k,&m,&q);
        int cnt=0,s=0;
        for(int i=0;i<k;i++){
            scanf("%s %d",a[i].name,&a[i].v);
            a[i].num=i;
        }
        //op o[150];
        for(int i=0;i<m;i++){
            scanf("%d%d",&o[i].t,&o[i].p);
        }
        sort(o,o+m,cmp);
        for(int i=0;i<m;i++){
            while(cnt<o[i].t){
                que.push(a[cnt++]);
            }
            while(o[i].p--&&!que.empty()){
                node temp=que.top();
                que.pop();
                strcpy(res[++s],temp.name);
            }
        }
        while(cnt<k){
            que.push(a[cnt++]);
        }
        while(!que.empty()){
            node temp=que.top();
            que.pop();
            strcpy(res[++s],temp.name);
        }
        while(q--){
            int n;
            scanf("%d",&n);
            printf("%s%c",res[n],q==0?'\n':' ');
        }
    }
    return 0;
}