Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2650    Accepted Submission(s): 722

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
题意:有k个人来,门先会开m次,每次开门的条件是假设当第a个人来时,可以放b个人进来。最后会把人全部放进来。查询第i个进来的是谁?
收获:题目没有说开门条件第a个人是按从小到大排序的,而且也没说a一定大于b;所以要考虑这种情况,保持程序的健壮性。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstdlib>
#include<string>
#include<map>
#include<stack> using namespace std;
#define ll long long
#define lson le,mid,num<<1
#define rson mid+1,ri,num<<1|1
#define stop1 puts("QAQ")
#define stop2 system("pause")
#define maxn 160000 struct node
{
char s[];
int val;
int id;
bool operator < (const node & a) const
{
if(val!=a.val)
return val < a.val;
return id>a.id;
} }peo[maxn];
struct kk
{
int x,y;
}tt[maxn]; int cmp(kk a, kk b)
{
return a.x < b.x;
}
char ans[maxn][]; int main()
{ int T;
scanf("%d",&T);
while(T--)
{ int k,m,q;
scanf("%d%d%d",&k,&m,&q);
for(int i=;i<=k;i++)
{
scanf("%s%d",peo[i].s,&peo[i].val);
peo[i].id=i;
} for(int i=;i<=m;++i)
scanf("%d%d",&tt[i].x,&tt[i].y); sort(tt+, tt+m+,cmp);
priority_queue<node> Q;
memset(ans, , sizeof ans); int sum=,pos=;
for(int i=;i<=m;i++)
{ for(int j = pos; j <=tt[i].x; j++)
{
Q.push(peo[j]);
pos++;
}
for(int j=;j<=tt[i].y;j++)
{
if(!Q.empty())
{node temp = Q.top();
Q.pop();
sum++;
strcpy(ans[sum],temp.s);
}
else
break;
}
} for(int i=pos;i<=k;i++)
{
Q.push(peo[i]);
}
while(!Q.empty())
{
node temp = Q.top();
Q.pop();
sum++;
strcpy(ans[sum],temp.s);
} int ss;
for(int i=;i<q;i++)
{
scanf("%d",&ss);
printf("%s ",ans[ss]);
}
if(q!=)
{
scanf("%d",&ss);
printf("%s\n",ans[ss]);
} }
return ;
}
 

HDU5437 Alisha’s Party (优先队列 + 模拟)的更多相关文章

  1. HDU5437 Alisha’s Party 优先队列

    点击打开链接 可能出现的问题: 1.当门外人数不足p人时没有判断队列非空,导致RE. 2.在m次开门之后最后进来到一批人没有入队. 3.给定的开门时间可能是打乱的,需要进行排序. #include&l ...

  2. Alisha’s Party (HDU5437)优先队列+模拟

    Alisha 举办聚会,会在一定朋友到达时打开门,并允许相应数量的朋友进入,带的礼物价值大的先进,最后一个人到达之后放外面的所有人进来.用优先队列模拟即可.需要定义朋友结构体,存储每个人的到达顺序以及 ...

  3. HDU 5437 Alisha’s Party (优先队列模拟)

    题意:邀请k个朋友,每个朋友带有礼物价值不一,m次开门,每次开门让一定人数p(如果门外人数少于p,全都进去)进来,当最后所有人都到了还会再开一次门,让还没进来的人进来,每次都是礼物价值高的人先进.最后 ...

  4. hdu 5437(优先队列模拟)

    Alisha’s Party Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  5. hdu5437 Alisha’s Party

    Problem Description Princess Alisha invites her friends to come to her birthday party. Each of her f ...

  6. Codeforces Round #318 (Div. 2) A Bear and Elections (优先队列模拟,水题)

    优先队列模拟一下就好. #include<bits/stdc++.h> using namespace std; priority_queue<int>q; int main( ...

  7. Alisha’s Party---hdu5437(模拟+优先队列)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5437 题意:公主有k个朋友来参加她的生日party,每个人都会带价值为v[i]的礼物过来,在所有人到齐 ...

  8. 优先队列 + 模拟 - HDU 5437 Alisha’s Party

    Alisha’s Party Problem's Link Mean: Alisha过生日,有k个朋友来参加聚会,由于空间有限,Alisha每次开门只能让p个人进来,而且带的礼物价值越高就越先进入. ...

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

    Problem Description Princess Alisha invites her friends to come to her birthday party. Each of her f ...

随机推荐

  1. Java应用开发的一条经验

    一旦为应用建立良好的基础设施, 后续的开发就会变得容易而快速.  这些基础设施包括: 1.   线程池的建立.配置: 在 JDK 并发库的基础上建立更适合于应用的并发使用接口: 2.   跨多数据源的 ...

  2. 第05讲- DDMS中logcat的使用

    第05讲 DDMS中logcat的使用 1.DDMS DDMS 的全称是Dalvik Debug Monitor Service,是 Android 开发环境中的Dalvik虚拟机调试监控服务.DDM ...

  3. java技术学习网址收藏

    Bootstrap:http://www.runoob.com/bootstrap/bootstrap-intro.html AngularJS : http://www.runoob.com/ang ...

  4. PC-CSS-分隔线

    单个标签实现分隔线: 点此查看实例展示 .demo_line_01{ padding: 0 20px 0; margin: 20px 0; line-height: 1px; border-left: ...

  5. javascript 执行环境,变量对象,作用域链

    前言 这几天在看<javascript高级程序设计>,看到执行环境和作用域链的时候,就有些模糊了.书中还是讲的不够具体. 通过上网查资料,特来总结,以备回顾和修正. 要讲的依次为: EC( ...

  6. jqGrid源代码分析(一)

    废话少说.先上grid.base.js 整体结构图 watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvc3B5MTk4ODEyMDE=/font/5a6L5L2 ...

  7. Lua内存泄漏应对方法[转]

    转自http://blog.csdn.net/xocoder/article/details/42685685 由于目前正在负责的项目是一个二次开发项目,而且留给我们的代码质量实在让人无力吐槽,所以遇 ...

  8. Linux 下提高make的编译效率

    Linux下安装程序,一般都通过包管理器安装,但是包管理器或软件商店里的软件往往不是最新版本的,安装最新版软件时通常是下载源代码进行编译. 编译安装源代码时就离不开make了,但是make是单线程的, ...

  9. [HeadFirst-HTMLCSS学习笔记][第十四章交互活动]

    表单 <form action="http://wickedlysmart.com/hfhtmlcss/contest.php" method="POST" ...

  10. xtrabackup备份恢复测试

    http://blog.chinaunix.net/uid-20682026-id-3319204.html