HDU 5437 Alisha’s Party（模拟）

Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1494 Accepted Submission(s): 401



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：一个公主要举行生日party，n个朋友给她来过生日并且每个人都带了礼物，由于房间不够所以公主想了一个主意她开m此门，每次开门放p个人进来，进门的顺序按照每个人带来礼物的价值来决定，价值高的人优先进入，如果价值相同先来的人先进。然后后面有k次询问，每次询问输入一个数字x，输出第x个进入房间的人的名字
ps：m次开门之后，如果门外还有人的话就等到当所有好友都来齐的时候进入，顺序还是按照礼物的价值大小。

模拟水题：

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
#include<math.h>

using namespace std;

struct node{
     int x;
     int y;
     char name[210];
}p[150100];

struct node1{
    int x1;
    int y1;
}ai[151000];

bool operator < (const node &a, const node &b)
{
    if(a.x == b.x){
        return a.y>b.y;
    }
    return a.x < b.x;
}

int cmp(const void *a,const void *b){
    struct node1 *aa = (node1*)a;
    struct node1 *bb = (node1*)b;
    return aa->x1 - bb->x1;
}

priority_queue<node>q;

int n,m,k;
char Name[150100][210];
int w[150001];

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1;i<=n;i++){
            scanf("%s%d",p[i].name,&p[i].x);
            p[i].y = i;
        }
        while(!q.empty()){
            q.pop();
        }
        int px = 0;
        int t = 1;
        for(int i=0;i<m;i++){
            scanf("%d%d",&ai[i].x1,&ai[i].y1);
        }
        qsort(ai,m,sizeof(ai[0]),cmp);
        struct node f;
        for(int ii=0;ii<m;ii++){
            int xx,yy;
            xx = ai[ii].x1;
            yy = ai[ii].y1;
            for(int i=xx;i>px;i--){
                q.push(p[i]);
            }
            while(yy){
                if(!q.empty()){
                    f = q.top();
                    q.pop();
                    strcpy(Name[t],f.name);
                    t++;
                }else{
                    break;
                }
                yy--;
            }
            px = xx;
        }
        for(int i=n;i>px;i--){
            q.push(p[i]);
        }
        while(!q.empty()){
            f = q.top();
            strcpy(Name[t],f.name);
            t++;
            q.pop();
        }
        for(int i=1;i<=k;i++){
            scanf("%d",&w[i]);
        }
        for(int i=1;i<=k;i++){
            if(i<k){
                printf("%s ",Name[w[i]]);
            }else{
                printf("%s",Name[w[i]]);
            }
        }
        printf("\n");
    }
    return 0;
}