time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Very soon Berland will hold a School Team Programming Olympiad. From each of the m Berland regions a team of two people is invited to participate in the olympiad. The qualifying contest to form teams was held and it was attended by n Berland students. There were at least two schoolboys participating from each of the m regions of Berland. The result of each of the participants of the qualifying competition is an integer score from 0 to 800 inclusive.

The team of each region is formed from two such members of the qualifying competition of the region, that none of them can be replaced by a schoolboy of the same region, not included in the team and who received a greater number of points. There may be a situation where a team of some region can not be formed uniquely, that is, there is more than one school team that meets the properties described above. In this case, the region needs to undertake an additional contest. The two teams in the region are considered to be different if there is at least one schoolboy who is included in one team and is not included in the other team. It is guaranteed that for each region at least two its representatives participated in the qualifying contest.

Your task is, given the results of the qualifying competition, to identify the team from each region, or to announce that in this region its formation requires additional contests.

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 10 000, n ≥ 2m) — the number of participants of the qualifying contest and the number of regions in Berland.

Next n lines contain the description of the participants of the qualifying contest in the following format: Surname (a string of length from 1 to 10 characters and consisting of large and small English letters), region number (integer from 1 to m) and the number of points scored by the participant (integer from 0 to 800, inclusive).

It is guaranteed that all surnames of all the participants are distinct and at least two people participated from each of the m regions. The surnames that only differ in letter cases, should be considered distinct.

Output

Print m lines. On the i-th line print the team of the i-th region — the surnames of the two team members in an arbitrary order, or a single character "?" (without the quotes) if you need to spend further qualifying contests in the region.

Examples
Input
5 2
Ivanov 1 763
Andreev 2 800
Petrov 1 595
Sidorov 1 790
Semenov 2 503
Output
Sidorov Ivanov
Andreev Semenov
Input
5 2
Ivanov 1 800
Andreev 2 763
Petrov 1 800
Sidorov 1 800
Semenov 2 503
Output
?
Andreev Semenov
Note

In the first sample region teams are uniquely determined.

In the second sample the team from region 2 is uniquely determined and the team from region 1 can have three teams: "Petrov"-"Sidorov", "Ivanov"-"Sidorov", "Ivanov" -"Petrov", so it is impossible to determine a team uniquely.

优先队列来做,想起来那个病人优先级那个题了。

 #include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = 1e4+;
struct node{
char s[];
int g;
friend bool operator < (node A,node B){
return A.g<B.g;
}
};
priority_queue<node> p[maxn];
void solve(){
int n,m;
scanf("%d%d", &n,&m);
for(int i = ; i<=n; i++){
int r;
node nod;
scanf("%s%d%d", nod.s,&r,&nod.g);
p[r].push(nod);
}
for(int i = ; i<=m; i++){
node nod1,nod2,nod3;
nod1 = p[i].top();
p[i].pop();
nod2 = p[i].top();
p[i].pop();
if(p[i].size()>){
nod3 = p[i].top();
p[i].pop();
if(nod2.g == nod3.g){
printf("?\n");
}
else{
printf("%s %s\n",nod1.s,nod2.s);
}
} else printf("%s %s\n",nod1.s,nod2.s);
}
}
int main()
{
solve();
return ;
}

卷珠帘

B. Qualifying Contest的更多相关文章

  1. Codeforces Round #346 (Div. 2) B Qualifying Contest

    B. Qualifying Contest 题目链接http://codeforces.com/contest/659/problem/B Description Very soon Berland ...

  2. Codeforces Round #346 (Div. 2) B. Qualifying Contest 水题

    B. Qualifying Contest 题目连接: http://www.codeforces.com/contest/659/problem/B Description Very soon Be ...

  3. codeforces 659B B. Qualifying Contest(水题+sort)

    题目链接: B. Qualifying Contest time limit per test 1 second memory limit per test 256 megabytes input s ...

  4. Codeforces 659B Qualifying Contest【模拟,读题】

    写这道题题解的目的就是纪念一下半个小时才读懂题...英文一多读一读就溜号... 读题时还时要静下心来... 题目链接: http://codeforces.com/contest/659/proble ...

  5. codeforces 659B Qualifying Contest

    题目链接:http://codeforces.com/problemset/problem/659/B 题意: n个人,m个区.给出n个人的姓名(保证不相同),属于的区域,所得分数.从每个区域中选出成 ...

  6. B. Qualifying Contest_排序

    B. Qualifying Contest time limit per test 1 second memory limit per test 256 megabytes input standar ...

  7. Codeforces 659 - A/B/C/D/E/F/G - (Undone)

    链接:https://codeforces.com/contest/659 A - Round House - [取模] AC代码: #include<bits/stdc++.h> usi ...

  8. codeforces659B

    Qualifying Contest CodeForces - 659B Very soon Berland will hold a School Team Programming Olympiad. ...

  9. Codeforces Round #346 (Div. 2) B题

    B. Qualifying Contest Very soon Berland will hold a School Team Programming Olympiad. From each of t ...

随机推荐

  1. iOS 导航栏去阴影

    if ([[[UIDevicecurrentDevice] systemVersion] floatValue] >= 6.0) { // 首先要判断版本号,否则在iOS 6 以下的版本会闪退 ...

  2. 关于Spring Security 3获取用户信息的问题

    标签: spring security 3标签获取用户信息 2013-01-05 10:40 5342人阅读 评论(0) 收藏 举报  分类: Spring(25) java(70) 前端(7)    ...

  3. JVM 设置

    按照基本回收策略分 引用计数(Reference Counting) 标记-清除(Mark-Sweep) 复制(Copying) 标记-整理(Mark-Compact) 按分区对待的方式分 增量收集( ...

  4. Ansible5:常用模块【转】

    根据zs官方的分类,将模块按功能分类为:云模块.命令模块.数据库模块.文件模块.资产模块.消息模块.监控模块.网络模块.通知模块.包管理模块.源码控制模块.系统模块.单元模块.web设施模块.wind ...

  5. Ansible1:简介与基本安装【转】

    Ansible是一个综合的强大的管理工具,他可以对多台主机安装操作系统,并为这些主机安装不同的应用程序,也可以通知指挥这些主机完成不同的任务.查看多台主机的各种信息的状态等,ansible都可以通过模 ...

  6. SSH自动断开连接的原因

    方法一: 用putty/SecureCRT连续3分钟左右没有输入, 就自动断开, 然后必须重新登陆, 很麻烦. 在网上查了很多资料, 发现原因有多种, 环境变量TMOUT引起,ClientAliveC ...

  7. JSP导出Excel后身份证后三位为0

    JSP导出Excel身份证号码超出Excel最大限制,用科学计数法表示,但后三位为0,修改方式: <style type="text/css">.txt    {    ...

  8. The magic behind configure, make, make install

    原文:https://robots.thoughtbot.com/the-magic-behind-configure-make-make-install#where-do-these-scripts ...

  9. 【转】ethtool 命令详解

    命令描述: ethtool 是用于查询及设置网卡参数的命令. 使用概要:ethtool ethx       //查询ethx网口基本设置,其中 x 是对应网卡的编号,如eth0.eth1等等etht ...

  10. 单元测试、自动化测试、接口测试过程中的Excel数据驱动(java实现)

    import java.io.FileInputStream;import java.io.InputStream;import java.util.HashMap;import java.util. ...