HDU5437 Alisha’s Party（优先队列+模拟）

Alisha’s Party

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

Total Submission(s): 4075    Accepted Submission(s): 1052

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个人带着价值v_i的礼物来，开m次门，每次在有t个人来的时候开门放进来p个人，所有人都来了之后再开一次门把剩下的人都放进来，每次带礼物价值高的人先进，价值相同先来先进，q次询问，询问第n个进来的人的名字。

分析：优先队列+模拟就可以了，只是注意m可以为0。

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define MAXN 150010

struct node
{
    int id,v;
    char name[220];
    bool operator < (const node &tmp) const
    {
        if(v == tmp.v) return id < tmp.id;
        return v > tmp.v;
    }
}p[MAXN];

struct opendoor
{
    int a,b;
    bool operator < (const opendoor &tmp) const
    {
        return a < tmp.a;
    }
}od[MAXN];

set<node> s;
int query[110];

int main()
{
    int T,n,m,q;
    scanf("%d",&T);
    while(T--)
    {
        s.clear();
        scanf("%d%d%d",&n,&m,&q);
        for(int i=1; i<=n; i++)
        {
            scanf("%s %d",p[i].name, &p[i].v);
            p[i].id = i;
        }
        for(int i=0; i<m; i++)
            scanf("%d%d",&od[i].a, &od[i].b);
        sort(od, od+m);
        int maxq;
        for(int i=0; i<q; i++)
        {
            scanf("%d",&query[i]);
            maxq = max(maxq, query[i]);
        }
        vector <int> ans;
        int cnt = 0;
        for(int i=1; i<=n&&ans.size()<maxq; i++)
        {
            s.insert(p[i]);
            while(od[cnt].a==i && cnt<m)
            {
                for(int j=0; j<od[cnt].b&&!s.empty()&&ans.size()<maxq; j++)
                {
                    ans.push_back(s.begin()->id);
                    s.erase(s.begin());
                }
                cnt++;
            }
        }
        while(!s.empty() && ans.size()<maxq)
        {
            ans.push_back(s.begin()->id);
            s.erase(s.begin());
        }
        for(int i=0; i<q; i++)
        {
            if(i) printf(" ");
            cout<<p[ans[query[i]-1]].name;
        }
        puts("");
    }
    return 0;
}
</span>