B. Destruction of a Tree
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., pn (0 ≤ pi ≤ n). If pi ≠ 0 there is an edge between vertices i and pi. 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.

题目大意:给你一棵树,只能删除度数为偶数的节点,节点删除后,与它相连的边也会删除。问你能否把所有点删除。

解题思路:只要你能想到,如果一棵树,有偶数条边,那么他一定能被删除完!或者说,一棵树有奇数个节点,那么他肯定能被删除完!因为,如果边为奇数,每次删除偶数条边,最后肯定剩奇数个边啊!如果边为偶数,每次删除偶数条边,最后肯定能删除完!所以基于这个思想,我们递归的删除点即可。从根节点开始,如果某一棵子树他的节点个数为偶数(加上当前节点就为奇数了),那么就深搜这颗子树,递归删除。

#include <bits/stdc++.h>
using namespace std; vector<int> ch[];
int sz[]; void getsize(int u, int pre)
{
sz[u] = ;
for (int i = ; i < ch[u].size(); ++i)
{
if (ch[u][i] != pre)
{
getsize(ch[u][i], u);
sz[u] += sz[ch[u][i]];
}
}
} void dfs(int u, int pre)
{
for (int i = ; i < ch[u].size(); i++)
{
if (ch[u][i] != pre)
{
if (sz[ch[u][i]] % == )
{
dfs(ch[u][i], u);
}
}
} printf("%d\n", u); for (int i = ; i < ch[u].size(); i++)
{
if (ch[u][i] != pre)
{
if (sz[ch[u][i]] % == )
{
dfs(ch[u][i], u);
}
}
}
} int main()
{ int N;
scanf("%d", &N);
int temp;
int root;
for (int i = ; i <= N; i++)
{
scanf("%d", &temp);
if (temp != )
{
ch[i].push_back(temp);
ch[temp].push_back(i);
}
else
{
root = i;
}
} if (N % == )
{
printf("NO\n");
}
else
{
getsize(root,-);
printf("YES\n");
dfs(root, -);
} return ;
}

我wa的代码,原因是,我的写法是每次找出边是偶数的,删掉,再继续,仔细思考找到了可以把我wa的样例:

先删1的话,会导致输出“NO”

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#include<queue>
#define ll unsigned long long
#define inf 0x3f3f3f3f
using namespace std;
vector<int>v[];
bool us[];
int in[];
queue<int>q;
int main()
{
int n;
cin>>n;
memset(us,,sizeof(us));
memset(in,,sizeof(in));
while(!q.empty()) q.pop();
for(int i=;i<=n;i++)
{
int x;
cin>>x;
if(x!=)
{
v[x].push_back (i);
v[i].push_back (x);
in[i]++;
in[x]++;
}
}
if(n%==) cout<<"NO"<<endl;
else
{
while()
{
int k=-;
for(int i=;i<=n;i++)
{
if(!us[i]&&in[i]%==)
{
k=i;
break;
}
}
if(k==-) break;
q.push(k);
us[k]=;
for(int j=;j<v[k].size ();j++)
{
int y=v[k][j];
if(us[y]) continue;
in[y]--;
}
}
bool f=;
for(int i=;i<=n;i++)
{
if(!us[i])
{
f=;
break;
}
}
if(!f) cout<<"NO";
else
{
cout<<"YES"<<endl;
while(!q.empty ())
{
cout<<q.front()<<endl;
q.pop();
}
}
}
return ;
}

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