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 ;
}

CodeForces - 963B Destruction of a Tree (dfs+思维题)的更多相关文章

  1. codeforces 963B Destruction of a Tree

    B. Destruction of a Tree time limit per test 1 second memory limit per test 256 megabytes input stan ...

  2. Codeforces 963B Destruction of a Tree 思维+dfs

    题目大意: 给出一棵树,每次只能摧毁有偶数个度的节点,摧毁该节点后所有该节点连着的边都摧毁,判断一棵树能否被摧毁,若能,按顺序输出摧毁的点,如果有多种顺序,输出一种即可 基本思路: 1)我一开始自然而 ...

  3. codeforces 812E Sagheer and Apple Tree(思维、nim博弈)

    codeforces 812E Sagheer and Apple Tree 题意 一棵带点权有根树,保证所有叶子节点到根的距离同奇偶. 每次可以选择一个点,把它的点权删除x,它的某个儿子的点权增加x ...

  4. Codeforces 878D - Magic Breeding(bitset,思维题)

    题面传送门 很容易发现一件事情,那就是数组的每一位都是独立的,但由于这题数组长度 \(n\) 很大,我们不能每次修改都枚举每一位更新其对答案的贡献,这样复杂度必炸无疑.但是这题有个显然的突破口,那就是 ...

  5. XJOI3363 树3/Codeforces 682C Alyona and the Tree(dfs)

    Alyona decided to go on a diet and went to the forest to get some apples. There she unexpectedly fou ...

  6. codeforces 29D Ant on the Tree (dfs,tree,最近公共祖先)

    D. Ant on the Tree time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  7. Codeforces Gym101246G:Revolutionary Roads(DFS+思维)

    http://codeforces.com/gym/101246/problem/G 题意:有一个n个点m条边的有向图,现在可以修改某一条有向边使得其为无向边,问修改哪些边可以使得修改后的强连通分量的 ...

  8. codeforces 799 D. Field expansion(dfs+思维剪枝)

    题目链接:http://codeforces.com/contest/799/problem/D 题意:给出h*w的矩阵,要求经过操作使得h*w的矩阵能够放下a*b的矩阵,操作为:将长或者宽*z[i] ...

  9. codeforces 682C Alyona and the Tree DFS

    这个题就是在dfs的过程中记录到根的前缀和,以及前缀和的最小值 #include <cstdio> #include <iostream> #include <ctime ...

随机推荐

  1. L3-015. 球队“食物链”(dfs)

    L3-015. 球队“食物链” 某国的足球联赛中有N支参赛球队,编号从1至N.联赛采用主客场双循环赛制,参赛球队两两之间在双方主场各赛一场. 联赛战罢,结果已经尘埃落定.此时,联赛主席突发奇想,希望从 ...

  2. python装饰器中functools.wraps的作用详解

    直接上代码看效果: # 定义一个最简单的装饰器 def user_login_data(f): def wrapper(*args, **kwargs): return f(*args, **kwar ...

  3. CSDN博客积分规则

    1.博客积分规则 博客积分是CSDN对用户努力的认可和奖励,也是衡量博客水平的重要标准.博客等级也将由博客积分唯一决定.积分规则具体如下: 每发布一篇原创或者翻译文章:可获得10分: 每发布一篇转载文 ...

  4. String.prototype.getParm

    String.prototype.getParms=function(name){ var reg = new RegExp('(^|&)' + name + '=([^&]*)(&a ...

  5. 解决h5py的FutureWarning问题

    h5py/__init__.py:: FutureWarning: Conversion of the second argument of issubdtype from `float` to `n ...

  6. HDU 数位dp

    模板http://www.cnblogs.com/jffifa/archive/2012/08/17/2644847.html 完全理解以后,我发现这种写法实在是太厉害了,简洁,优美,可以回避很多细节 ...

  7. HTML中Div、span、label标签的区别

    div与span 大家在初学div+css布局时,有很多困惑,在div与span的使用过程没觉得有一定的”章法”,觉得两个区别不大,在w3c的关于div和span的定义:div作为分割文档结构自然使它 ...

  8. UISegmentedControl字体大小,颜色,选中颜色,左边椭圆,右边直线的Button 解决之iOS开发之分段控制器UISegmentedControl

        NSArray *segmentedArray = [NSArrayarrayWithObjects:STR(@"Mynews"),STR(@"Systemmes ...

  9. boost split字符串

    boost split string , which is very convenience #include <string> #include <iostream> #in ...

  10. python 两种导入的区别

    参考文献: 法1:import module1 使用:这样你调用时就得:module1.funct(),funct2(),... 法2:from module1 import funct() 使用就直 ...