HDU 5437 Alisha’s Party
来源:互联网 发布:三季稻知乎 编辑:程序博客网 时间:2024/05/16 18:14
Alisha’s Party
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2507 Accepted Submission(s): 678
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 letp 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 queryn Please tell Alisha who the n−th person to enter her castle is.
Each time when Alisha opens the door, she can decide to let
If there are two friends who bring gifts of the same value, then the one who comes first should enter first. Given a query
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 numbersk,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 .
Thei−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 followingm 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 containq 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 containingn>10000 .
In each test case, the first line contains three numbers
The
Each of the following
The last line of each test case will contain
Note: there will be at most two test cases containing
Output
For each test case, output the corresponding name of Alisha’s query, separated by a space.
Sample Input
15 2 3Sorey 3Rose 3Maltran 3Lailah 5Mikleo 61 14 21 2 3
Sample Output
Sorey Lailah Rose
Source
2015 ACM/ICPC Asia Regional Changchun Online
Recommend
hujie | We have carefully selected several similar problems for you: 5449 5448 5447 5446 5445
ACcode:
#include <map>#include <queue>#include <cmath>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#include <string>#define rd(x) scanf("%d",&x)#define rd2(x,y) scanf("%d%d",&x,&y)#define rds(x) scanf("%s",x)#define ll long long int#define maxn 150000+1#define mod 1000000007#define INF 0x3f3f3f3f //int×î′óÖμ#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)#define MT(x,i) memset(x,i,sizeof(x))#define PI acos(-1.0)#define E exp(1)using namespace std;struct node{ int value; int time; friend bool operator <(node a,node b){ if(a.value==b.value) return a.time>b.time; else return a.value<b.value; }}my;char name[maxn][201];int v[maxn];int t[maxn];int data[maxn];int ti[maxn];int main(){ int loop,n,m,q; rd(loop); while(loop--){ memset(ti,0,sizeof(ti)); rd2(n,m);rd(q); FOR(i,1,n){ rds(name[i]); rd(v[i]); } priority_queue<node>que; int cnt=0; for(int i=1,x,y;i<=m;i++){ rd2(x,y); ti[x]=y; } FOR(i,1,n){ my.time=i; my.value=v[i]; que.push(my); if(i==n){ while(!que.empty()){ my=que.top(); data[++cnt]=my.time; que.pop(); } break; } FOR(j,1,ti[i]){ if(que.empty()) break; my=que.top(); que.pop(); data[++cnt]=my.time; } } int t; FOR(i,1,q){ rd(t); printf("%s",name[data[t]]); if(i==q) putchar('\n'); else putchar(' '); } } return 0;}/**15 2 3Sorey 3Rose 3Maltran 3Lailah 5Mikleo 61 14 21 2 3**/
0 0
- hdu 5437 Alisha’s Party
- hdu 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- HDU-5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- hdu 5437(Alisha’s Party)
- HDU 5437 Alisha’s Party
- HDU 5437 Alisha’s Party
- Alisha’s Party HDU
- HDU 5437 Alisha’s Party 优先队列
- hdu 5437 Alisha’s Party 优先队列
- HDU 5437 Alisha’s Party(模拟)
- hdu 5437Alisha’s Party(优先队列)
- hdu 5437 Alisha’s Party (优先队列)
- hdu 5437 Alisha’s Party 优先队列
- [HDU 5437]Alisha’s Party[模拟]
- utf-8 unicode 各种编码的区别与联系
- Selenium2(JAVA) Web自动化测试实战 电子书百度阅读正式上架 欢迎试读购买
- PAT Basic 1034 有理数四则运算(20)
- DataBase
- poj 2377 最小生成树(kruskal算法)
- HDU 5437 Alisha’s Party
- hdu 5024 Wang Xifeng's Little Plot(搜索)
- 猜商品价格
- ios开发-新浪微博12-(标题按钮的箭头上下翻转)
- 安卓中异步任务实现带下载进度的根据图片地址下载图片
- Python学习资料与博客推荐
- Yii2.0 邮件配置
- PHP中文乱码的三个原因及解决方法
- 2199 Can you solve this equation?