CodeForces - 1255C

Bob is an avid fan of the video game “League of Leesins”, and today he celebrates as the League of Leesins World Championship comes to an end!

The tournament consisted of n (n≥5) teams around the world. Before the tournament starts, Bob has made a prediction of the rankings of each team, from 1-st to n-th. After the final, he compared the prediction with the actual result and found out that the i-th team according to his prediction ended up at the pi-th position (1≤pi≤n, all pi are unique). In other words, p is a permutation of 1,2,…,n.

As Bob’s favorite League player is the famous “3ga”, he decided to write down every 3 consecutive elements of the permutation p. Formally, Bob created an array q of n−2 triples, where qi=(pi,pi+1,pi+2) for each 1≤i≤n−2. Bob was very proud of his array, so he showed it to his friend Alice.

After learning of Bob’s array, Alice declared that she could retrieve the permutation p even if Bob rearranges the elements of q and the elements within each triple. Of course, Bob did not believe in such magic, so he did just the same as above to see Alice’s respond.

For example, if n=5 and p=[1,4,2,3,5], then the original array q will be [(1,4,2),(4,2,3),(2,3,5)]. Bob can then rearrange the numbers within each triple and the positions of the triples to get [(4,3,2),(2,3,5),(4,1,2)]. Note that [(1,4,2),(4,2,2),(3,3,5)] is not a valid rearrangement of q, as Bob is not allowed to swap numbers belong to different triples.

As Alice’s friend, you know for sure that Alice was just trying to show off, so you decided to save her some face by giving her any permutation p that is consistent with the array q she was given.

Input
The first line contains a single integer n (5≤n≤105) — the size of permutation p.

The i-th of the next n−2 lines contains 3 integers qi,1, qi,2, qi,3 (1≤qi,j≤n) — the elements of the i-th triple of the rearranged (shuffled) array qi, in random order. Remember, that the numbers within each triple can be rearranged and also the positions of the triples can be rearranged.

It is guaranteed that there is at least one permutation p that is consistent with the input.

Output
Print n distinct integers p1,p2,…,pn (1≤pi≤n) such that p is consistent with array q.

If there are multiple answers, print any.
**Example
Input
5
4 3 2
2 3 5
4 1 2
Output
1 4 2 3 5 **
题意：
给定n-2个三元序列，每组序列代表的是三个人的顺序，然后根据给定的n-2行序列，推出n个选手的排名。
在所有序列中只出现一次的数字肯定是第一名或者最后一名，然后就可以先判断出第一名，与第一名在同一序列当中的两个人就是第二和第三名，第二名只会在两个序列中出现，因此，第二名第三名也判断出来了，前三名判断出来后，标记一下（代表已经确定了排名），然后再根据第二和第三名，判断后面的排名即可。

#include
#include
#include
using namespace std;
const int N = 5e5 + 500;
vector<int>an[N];
bool vis[N];
int main() {
	int n;
	scanf("%d",&n);
	for(int i=0; i<n-2; ++i) {
		int a,b,c;
		scanf("%d %d %d",&a,&b,&c);
		an[a].push_back(b);
		an[a].push_back(c);
		an[b].push_back(a);
		an[b].push_back(c);
		an[c].push_back(a);
		an[c].push_back(b);
	}
	int be,en,cut;
	for(int i=1; i<=n; ++i) {
		if(an[i].size()==2) {
			cut=i;
			be=an[cut][0];
			en=an[cut][1];
			if(an[be].size()>an[en].size())swap(be,en);
			printf("%d %d %d",cut,be,en);
			vis[cut]=true;
			vis[be]=true;
			vis[en]=true;
			break;
		}
	}
	for(int i=4; i<=n; ++i) {
		for(int j=0; j<an[be].size(); ++j) {
			if(!vis[an[be][j]]) {
				printf(" %d",an[be][j]);
				vis[an[be][j]]=true;
				cut=en;
				en=an[be][j];
				be=cut;
				break;
			}
		}
	}
	return 0;
}