F. Tree Factory
Bytelandian Tree Factory produces trees for all kinds of industrial applications. You have been tasked with optimizing the production of a certain type of tree for an especially large and important order.
The tree in question is a rooted tree with n vertices labelled with distinct integers from 0 to n−1. The vertex labelled 0 is the root of the tree, and for any non-root vertex v the label of its parent p(v) is less than the label of v.
All trees at the factory are made from bamboo blanks. A bamboo is a rooted tree such that each vertex has exactly one child, except for a single leaf vertex with no children. The vertices of a bamboo blank can be labelled arbitrarily before its processing is started.
To process a bamboo into another tree a single type of operation can be made: choose an arbitrary non-root vertex v such that its parent p(v) is not a root either. The operation consists of changing the parent of v to its parent's parent p(p(v)). Note that parents of all other vertices remain unchanged, in particular, the subtree of v does not change.
Efficiency is crucial, hence you have to minimize the number of operations to make the desired tree from a bamboo blank. Construct any optimal sequence of operations to produce the desired tree.
Note that the labelling of the resulting tree has to coincide with the labelling of the desired tree. Formally, the labels of the roots have to be equal, and for non-root vertices with the same label the labels of their parents should be the same.
It is guaranteed that for any test present in this problem an answer exists, and further, an optimal sequence contains at most 106 operations. Note that any hack that does not meet these conditions will be invalid.
Input
The first line contains a single integer n — the number of vertices in the tree (2≤n≤105).
The second line contains n−1 integers p(1),…,p(n−1) — indices of parent vertices of 1,…,n−1 respectively (0≤p(i)<i).
Output
In the first line, print n distinct integers id1,…,idn — the initial labelling of the bamboo blank starting from the root vertex (0≤idi<n).
In the second line, print a single integer k — the number of operations in your sequence (0≤k≤106).
In the third line print k integers v1,…,vk describing operations in order. The i-th operation consists of changing p(vi) to p(p(vi)). Each operation should be valid, i.e. neither vi nor p(vi) can be the root of the tree at the moment.
Examples
input
5
0 0 1 1
output
0 2 1 4 3
2
1 3
input
4
0 1 2
output
0 1 2 3
0
题意
现在有一个链，然后每次操作可以使得fa[x]=fa[fa[x]]，就是使得自己的爷爷变成自己的父亲。然后操作若干次，就会变成一颗树。
现在题目给你一个根为0的树，说这个是由一条链操作得到的，让你还原这条链的样子，并输出方案，要求操作次数最少。
题解
我们把操作转换一下，其实就是使得一个子树的父亲，变成自己的兄弟。
我们手动模拟一下，可以发现，树链的长度是一定的，我们如果让深度大于等于他的子树插进去的话，那么深度肯定会+1的，于是我们就每次插入深度大于等于他的就好了。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
int fa[maxn],mx[maxn],n;
vector<int>son[maxn];
vector<int>order;
bool cmp(int x,int y){
return mx[x]<mx[y];
}
void dfs1(int x){
for(int i=0;i<son[x].size();i++){
dfs1(son[x][i]);
mx[x]=max(mx[x],mx[son[x][i]]+1);
}
sort(son[x].begin(),son[x].end(),cmp);
}
void dfs2(int x){
cout<<x<<" ";
for(int i=0;i<son[x].size();i++){
dfs2(son[x][i]);
if(i==0)continue;
for(int j=0;j<=mx[son[x][i-1]];j++){
order.push_back(son[x][i]);
}
}
}
int main(){
scanf("%d",&n);
for(int i=1;i<n;i++){
scanf("%d",&fa[i]);
son[fa[i]].push_back(i);
}
dfs1(0);
dfs2(0);
cout<<endl;
cout<<order.size()<<endl;
for(int i=0;i<order.size();i++){
cout<<order[i]<<" ";
}
cout<<endl;
}