We are committed to the well being of all participants. Therefore, instead of the problem, we suggest you enjoy a piece of cake.
Uh oh. Somebody cut the cake. We told them to wait for you, but they did it anyway. There is still some left, though, if you hurry back. Of course, before you taste the cake, you thought about how the cake was cut.
It is known that the cake was originally a regular nn-sided polygon, each vertex of which had a unique number from 11 to nn. The vertices were numbered in random order.
Each piece of the cake is a triangle. The cake was cut into n−2n−2 pieces as follows: each time one cut was made with a knife (from one vertex to another) such that exactly one triangular piece was separated from the current cake, and the rest continued to be a convex polygon. In other words, each time three consecutive vertices of the polygon were selected and the corresponding triangle was cut off.
A possible process of cutting the cake is presented in the picture below.
Example of 6-sided cake slicing.
You are given a set of n−2n−2 triangular pieces in random order. The vertices of each piece are given in random order — clockwise or counterclockwise. Each piece is defined by three numbers — the numbers of the corresponding nn-sided cake vertices.
For example, for the situation in the picture above, you could be given a set of pieces: [3,6,5],[5,2,4],[5,4,6],[6,3,1][3,6,5].
You are interested in two questions.
What was the enumeration of the nn-sided cake vertices?
In what order were the pieces cut?
Formally, you have to find two permutations p1,p2,…,pnp1,p2,…,pn (1≤pi≤n1≤pi≤n) and q1,q2,…,qn−2q1,q2,…,qn−2 (1≤qi≤n−21≤qi≤n−2) such that if the cake vertices are numbered with the numbers p1,p2,…,pnp1,p2,…,pn in order clockwise or counterclockwise, then when cutting pieces of the cake in the order q1,q2,…,qn−2q1,q2,…,qn−2 always cuts off a triangular piece so that the remaining part forms one convex polygon.
For example, in the picture above the answer permutations could be: p=[2,4,6,1,3,5]p=[2,4,6,1,3,5] (or any of its cyclic shifts, or its reversal and after that any cyclic shift) and q=[2,4,1,3]q=[2,4,1,3].
Write a program that, based on the given triangular pieces, finds any suitable permutations pp and qq.
Input
The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases. Then there are tt independent sets of input data.
The first line of each set consists of a single integer nn (3≤n≤1053≤n≤105) — the number of vertices in the cake.
The following n−2n−2 lines describe the numbers of the pieces vertices: each line consists of three different integers a,b,ca,b,c (1≤a,b,c≤n1≤a,b,c≤n) — the numbers of the pieces vertices of cake given in random order. The pieces are given in random order.
It is guaranteed that the answer to each of the tests exists. It is also guaranteed that the sum of nn for all test cases does not exceed 105105.
Output
Print 2t2t lines — answers to given tt test cases in the order in which they are written in the input. Each answer should consist of 22 lines.
In the first line of an answer on a test case print nn distinct numbers p1,p2,…,pnp1,p2,…,pn(1≤pi≤n1≤pi≤n) — the numbers of the cake vertices in clockwise or counterclockwise order.
In the second line of an answer on a test case print n−2n−2 distinct numbers q1,q2,…,qn−2q1,q2,…,qn−2(1≤qi≤n−21≤qi≤n−2) — the order of cutting pieces of the cake. The number of a piece of the cake corresponds to its number in the input.
If there are several answers, print any. It is guaranteed that the answer to each of the tests exists.
Example
Input
3
6
3 6 5
5 2 4
5 4 6
6 3 1
6
2 5 6
2 5 1
4 1 2
1 3 5
3
1 2 3
Output
1 6 4 2 5 3
4 2 3 1
1 4 2 6 5 3
3 4 2 1
1 3 2
1
题意：给定一个由n边形生成的n-2个三角形各个顶点，求出这n边形的编号顺序以及每个三角形出现的顺序。
思路：首先切割的三角形，肯定有一个顶点只为这一个三角形所有，即这个点的度数为1。这样的话我们就按着拓扑排序的思想去bfs，这样就可以将第几次出现哪个三角形求出来了。对于n边形的顶点顺序，三角形的三条边，如果是n边形的边的话，那么它肯定是出现过一次且仅有一次！！！如果出现多次肯定不是n边形的边，这样的话我们把所有出现过一次边找出来，然后dfs一遍就可以找到n边形的顺序。
代码如下：
#include<bits/stdc++.h>
using namespace std;
const int maxx=1e5+100;
struct edge{
int next;
int to;
}e[maxx<<1];
struct node{
int a[4];
void Sort()
{
sort(a+1,a+1+3);
}
}p[maxx];
int head[maxx<<1],vis[maxx],deg[maxx],ans[maxx],mak[maxx];
vector<int> tt;
int n,tot=0;
inline void add(int u,int v)
{
e[tot].to=v,e[tot].next=head[u],head[u]=tot++;
}
inline void dfs(int u)
{
tt.push_back(u);
vis[u]=1;
for(int i=head[u];i!=-1;i=e[i].next)
{
int to=e[i].to;
if(vis[to]) continue;
dfs(to);
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
map<pair<int,int>,int> mp;
mp.clear();tt.clear();
vector<int> in[n+1];
for(int i=0;i<=n;i++) vis[i]=0,deg[i]=0,head[i]=-1,mak[i]=0;
for(int i=1;i<=n-2;i++)
{
for(int j=1;j<=3;j++) scanf("%d",&p[i].a[j]),deg[p[i].a[j]]++,in[p[i].a[j]].push_back(i);
p[i].Sort();
mp[make_pair(p[i].a[1],p[i].a[2])]^=1;
mp[make_pair(p[i].a[1],p[i].a[3])]^=1;
mp[make_pair(p[i].a[2],p[i].a[3])]^=1;
}
tot=0;
for(int i=1;i<=n-2;i++)
{
if(mp[make_pair(p[i].a[1],p[i].a[2])]==1) add(p[i].a[1],p[i].a[2]),add(p[i].a[2],p[i].a[1]);
if(mp[make_pair(p[i].a[1],p[i].a[3])]==1) add(p[i].a[1],p[i].a[3]),add(p[i].a[3],p[i].a[1]);
if(mp[make_pair(p[i].a[2],p[i].a[3])]==1) add(p[i].a[2],p[i].a[3]),add(p[i].a[3],p[i].a[2]);
}
dfs(1);
for(int i=0;i<tt.size();i++) cout<<tt[i]<<" ";
cout<<endl;
queue<int> q;
for(int i=1;i<=n;i++) if(deg[i]==1) q.push(i);
int cnt=0;
while(q.size())
{
int xx=q.front();
q.pop();
for(int k=0;k<in[xx].size();k++)
{
int y=in[xx][k];
if(mak[y]) continue;
ans[++cnt]=y;
mak[y]=1;
for(int i=1;i<=3;i++)
{
if(deg[p[y].a[i]]-1==1) q.push(p[y].a[i]);
--deg[p[y].a[i]];
}
}
}
for(int i=1;i<=n-2;i++) cout<<ans[i]<<" ";
cout<<endl;
}
return 0;
}
努力加油a啊，(o)/~
