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. Git 暂存区的概念

    工作区:我们在电脑里面能看到的目录,也就是我们用git init 命令初始化的那个目录.里面包含要添加文件和需要提交的文件,在这个目录下的文件,修改和变更,我们的git都能感知的到. 版本库:工作区有 ...

  2. mvc页面中,显示自定义时间格式

    1.在model中,遇到datetime格式的字段,用string来表示 model中 /// <summary> /// 开始时间 /// </summary> [Displ ...

  3. 缩放系列(二):所有子控件也随着缩放、手势缩放、多点触控layout

    下面是一个功能强大的改造的例子: 可以实现以下需求: 1.两个手指进行缩放布局 2.所有子控件也随着缩放, 3.子控件该有的功能不能丢失(像button有可被点击的功能,缩放后不能丢失该功能) 运行效 ...

  4. max_%_connection参数详解

    对于连接数的设置,show variables里有三个参数可以对它进行控制,max_connections与max_user_connections以及max_connect_errors.下面对这三 ...

  5. C++中的函数指针和函数对象总结

    篇一.函数指针函数指针:是指向函数的指针变量,在C编译时,每一个函数都有一个入口地址,那么这个指向这个函数的函数指针便指向这个地址.函数指针的用途是很大的,主要有两个作用:用作调用函数和做函数的参数. ...

  6. Swift --> Map & FlatMap

    转载自:https://segmentfault.com/a/1190000004050907 Map map函数能够被数组调用,它接受一个闭包作为参数,作用于数组中的每个元素.闭包返回一个变换后的元 ...

  7. nand驱动移植

    首先下载nand flash驱动 s3c_nand.c ,此文件包含着nand flash驱动具体的实现,将其复制到drivers/mtd/nand下: s3c_nand.c 下载地址 s3c_nan ...

  8. HTML一些小细节

    这里主要是记录一些看起来不重要,但是其实作用不小或者使用起来某种情况下好用的东西,又或者是重要的但容易忽略的基础知识 1. HTML5之后的第一个标签是<!DOCTYPE html> 这个 ...

  9. HDU-1072 Nightmare (bfs+贪心)

    Nightmare Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Sub ...

  10. android viewpager 深究

    参考: http://blog.csdn.net/xushuaic/article/details/42638311 GitHub上相关的ViewPager动画的项目 https://github.c ...