Alisha’s Party
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4389 Accepted Submission(s): 1121
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.
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.
5 2 3
Sorey 3
Rose 3
Maltran 3
Lailah 5
Mikleo 6
1 1
4 2
1 2 3
题意:有n个人要来参加聚会，主人会根据来的人所带来的礼物的价值让人进入房间，有m组操作，代表到第ti个人到场的时候主人会让pi个人进入房间，当人全部来齐之后主人会让所有没有进入过房间的人进入房间。然后给出q组询问,每一组询问aski代表第aski号进入房间的人的姓名。
#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
#include<queue>
using namespace std;
const int N = ;
int ans[N];
struct Node
{
char name[];
int time,v;
} node[N];
struct Node1
{
int t,p;
} node1[N];
bool operator <(Node a,Node b)
{
if(a.v==b.v) return a.time>b.time;
return a.v<b.v;
}
priority_queue<Node> qu;
int cmp(Node1 a,Node1 b)
{
return a.t<b.t;
}
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--)
{
int n,m,q;
scanf("%d%d%d",&n,&m,&q);
for(int i=; i<=n; i++)
{
scanf("%s%d",node[i].name,&node[i].v);
node[i].time=i;
}
int id = ,cnt=;
for(int i=; i<=m; i++)
{
scanf("%d%d",&node1[i].t,&node1[i].p);
}
sort(node1+,node1++m,cmp); ///之前先排序
for(int i=; i<=m; i++)
{
for(int j=id; j<=node1[i].t; j++) ///从第id号人开始进入房间
{
qu.push(node[id++]);
}
for(int j=; j<=node1[i].p; j++)
{
if(!qu.empty())
{
Node no = qu.top();
ans[cnt++] = no.time;
qu.pop();
}
}
}
if(id<=n) ///如果还有人没进去，依次进入
for(int i=id; i<=n; i++) qu.push(node[i]);
while(!qu.empty())
{
Node no = qu.top();
ans[cnt++] = no.time;
qu.pop();
}
for(int i=; i<q; i++)
{
int a;
scanf("%d",&a);
if(i<q-)printf("%s ",node[ans[a]].name);
else printf("%s\n",node[ans[a]].name);
}
}
return ;
}