HDU 5437 Alisha’s Party(优先队列+模拟)——2015 ACM/ICPC Asia Regional Changchun Online

Alisha’s Party

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

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：Princess Alisha邀请k个朋友来参加聚会，但是由于空间有限，Princess Alisha会打开m次门，每次会在第ti个人到的时候打开门，按要求(所带礼物价值高的先进，价值相同的先到先进)邀请等待中的pi个人进房间。等到所有人都到了之后，门会再次打开，剩下没有进房间的朋友会按照之前的方法依次进房间。q次询问，每次询问问第ni个进房间的人是谁

解题思路：此题坑的地方应该算是没告诉我们t是随机给出的，并不是按递增(或非递减)的顺序给出的，我们只需利用优先队列进行模拟就可以了，当然优先队列的优先级需要改一下，可以参考STL 优先队列 定义 优先级

另外对于q次询问，我们可以将结果记录下来，这样可以降低时间复杂度，毕竟题目给的内存限制足够大

#include<stdio.h>
#include<queue>
#include<algorithm>
#include<string.h>
using namespace std;
struct node
{    
    int v,id;
    char n[205];    
    bool operator < (const node &a) const   
    {    
        if(v!=a.v)
               return v<a.v;//最大值优先 
        return id>a.id;//最小值优先   
    }    
}s[150005];
struct person
{
    int t,p;
}u[150005];
bool cmp(person x,person y)
{
    return x.t<y.t;
}
int f_min(int a,int b)
{
    return a<b?a:b;
}
int main()
{
    int i,j,l,T,k,m,q,n,sum;
    scanf("%d",&T);
    while(T--)
    {
        priority_queue<node> d;
        scanf("%d%d%d",&k,&m,&q);
        for(i=0;i<k;i++)
        {
            scanf("%s%d",s[i].n,&s[i].v);
            s[i].id=i;
        }
        for(i=0;i<m;i++)
            scanf("%d%d",&u[i].t,&u[i].p);
        sort(u,u+m,cmp);
        for(sum=i=j=l=0;i<m;i++)
        {
            sum=f_min(sum+u[i].p,u[i].t);
            while(j<u[i].t&&j<k)
                d.push(s[j++]);
            while(l<sum&&!d.empty()&&l<k)
            {//printf("%s",d.top().n);
                strcpy(s[l++].n,d.top().n);
                d.pop();
            }
        }
        while(j<k)
            d.push(s[j++]);
        while(l<k&&!d.empty())
        {    
            strcpy(s[l++].n,d.top().n);
            d.pop();
        }
        for(i=0;i<q;i++)
        {
            scanf("%d",&n);
            printf("%s%c",s[n-1].n,i==q-1?'\n':' ');
        }
    }
    return 0;
}

菜鸟成长记