Destruction of a Tree

最新推荐文章于 2018-08-15 09:46:09 发布

原创最新推荐文章于 2018-08-15 09:46:09 发布 · 205 阅读

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本文介绍了一个算法问题，即如何在给定的树形结构中，按照特定条件摧毁所有顶点。该问题要求从一个具有偶数度的顶点开始，逐步移除符合条件的顶点及其相连边，直到树中不再存在任何顶点。文章提供了完整的实现代码，并通过两个示例展示了算法的工作过程。

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You are given a tree (a graph with n vertices and n - 1 edges in which it's possible to reach any vertex from any other vertex using only its edges).

A vertex can be destroyed if this vertex has even degree. If you destroy a vertex, all edges connected to it are also deleted.

Destroy all vertices in the given tree or determine that it is impossible.

Input

The first line contains integer n (1 ≤ n ≤ 2·10⁵) — number of vertices in a tree.

The second line contains n integers p₁, p₂, ..., p_n (0 ≤ p_i ≤ n). If p_i ≠ 0 there is an edge between vertices i and p_i. It is guaranteed that the given graph is a tree.

Output

If it's possible to destroy all vertices, print "YES" (without quotes), otherwise print "NO" (without quotes).

If it's possible to destroy all vertices, in the next n lines print the indices of the vertices in order you destroy them. If there are multiple correct answers, print any.

Examples

Input

Copy

5
0 1 2 1 2

Output

Copy

YES
1
2
3
5
4

Input

Copy

4
0 1 2 3

Output

Copy

NO

Note

In the first example at first you have to remove the vertex with index 1 (after that, the edges (1, 2) and (1, 4) are removed), then the vertex with index 2 (and edges (2, 3) and (2, 5) are removed). After that there are no edges in the tree, so you can remove remaining vertices in any order.

$\text{[math]}$

#include<bits/stdc++.h>
using namespace std;

const int N=2e5+10;
int p[N];
int d[N];

vector<int> e[N];
vector<int> ord;

void dfs(int v){
    for(int u:e[v])
        dfs(u);
    ord.push_back(v);
}

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    //freopen("in.txt","r",stdin);
    int n;
    cin>>n;
    int root=0;

    for(int i=1;i<=n;i++){
        cin>>p[i];
        ++d[p[i]];
        if(p[i]){
            d[i]++;
        }else{
            root=i;
        }
        e[p[i]].push_back(i);
    }
    dfs(root);

    vector<int> e;
    vector<int> ans;
    for(int i:ord){
        if(d[i]%2){
            if(i==root){
                puts("NO");return 0;
            }
            e.push_back(i);
        }else{
            ans.push_back(i);
            d[p[i]]--;
        }
    }
    reverse(e.begin(),e.end());
    ans.insert(ans.end(),e.begin(),e.end());
    cout<<"YES\n";
    //puts("YES");
    for(int x:ans){
        cout<<x<<endl;
    }
}