B. Qualifying Contest 结构体模拟

本文介绍了一场在Berland举行的学校团队编程奥林匹克竞赛的选拔过程。任务是在每个地区选出得分最高的两名参赛者组成团队,若出现分数相同导致无法唯一确定成员时,则需要额外的比赛来决定。文章详细解释了输入输出格式及示例。

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B. Qualifying Contest
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 0001 ≤ m ≤ 10 000n ≥ 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.


题意:n,m,n代表n个人,m代表m个队伍,然后,求出每个队伍的前两个分值最高的名字,比如第一个样例,两个队伍1和2,1分值最高的是790,先输出Sidorov,然后第二高的是763,再输出Ivanov,如果满足有三个最大值相同或者满足有一个最大值,和两个次大值相同,则输出‘?’,每个队伍的人数多于2,;
思路:直接结构体模拟,利用数组下标记录;
#include <stdio.h>
#include <algorithm>
using namespace std;
struct node {
	char c[25];
	int n;
	int m;
}s[100005];
struct node1{
	int a[805];
}s1[10005];
bool cmp(node a, node b) {
	if(a.n != b.n) 
		return a.n<b.n;
	else
		return a.m>b.m;
}
int main() {
	int n , m;
	scanf("%d%d",&n,&m);
	for(int i = 0; i < n; i++) {
		scanf("%s",s[i].c);
		scanf("%d",&s[i].n);
		scanf("%d",&s[i].m);
	}
	std::sort(s,s+n,cmp);
	for(int i = 0; i < n; i++) {
		s1[s[i].n].a[s[i].m]++;
	}
	int mm = 1 ,t = 0 ,k = 0 ,j;
	for(int i = 0; i < n; i++) {
		if(mm == s[i].n) {
			if(s1[mm].a[s[i].m] == 2) {
				printf("%s %s\n",s[i].c,s[i + 1].c);
				mm++;
			}
			else if(s1[mm].a[s[i].m] == 1 && s1[mm].a[s[i + 1].m] == 1) {
				printf("%s %s\n",s[i].c,s[i + 1].c);
				mm++;
			}
			else {
				printf("?\n");
				mm++;
			}
		}
	}
	return 0;
}


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