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
50 1 2 1 2
output
Copy
YES12354
input
Copy
40 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[200005];
int sz[200005];
void getsize(int u, int pre)
{
sz[u] = 1;
for (int i = 0; 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 = 0; i < ch[u].size(); i++)
{
if (ch[u][i] != pre)
{
if (sz[ch[u][i]] % 2 == 0)
{
dfs(ch[u][i], u);
}
}
}
printf("%d\n", u);
for (int i = 0; i < ch[u].size(); i++)
{
if (ch[u][i] != pre)
{
if (sz[ch[u][i]] % 2 == 1)
{
dfs(ch[u][i], u);
}
}
}
}
int main()
{
int N;
scanf("%d", &N);
int temp;
int root;
for (int i = 1; i <= N; i++)
{
scanf("%d", &temp);
if (temp != 0)
{
ch[i].push_back(temp);
ch[temp].push_back(i);
}
else
{
root = i;
}
}
if (N % 2 == 0)
{
printf("NO\n");
}
else
{
getsize(root,-1);
printf("YES\n");
dfs(root, -1);
}
return 0;
}
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[200005];
bool us[200005];
int in[200005];
queue<int>q;
int main()
{
int n;
cin>>n;
memset(us,0,sizeof(us));
memset(in,0,sizeof(in));
while(!q.empty()) q.pop();
for(int i=1;i<=n;i++)
{
int x;
cin>>x;
if(x!=0)
{
v[x].push_back (i);
v[i].push_back (x);
in[i]++;
in[x]++;
}
}
if(n%2==0) cout<<"NO"<<endl;
else
{
while(1)
{
int k=-1;
for(int i=1;i<=n;i++)
{
if(!us[i]&&in[i]%2==0)
{
k=i;
break;
}
}
if(k==-1) break;
q.push(k);
us[k]=1;
for(int j=0;j<v[k].size ();j++)
{
int y=v[k][j];
if(us[y]) continue;
in[y]--;
}
}
bool f=1;
for(int i=1;i<=n;i++)
{
if(!us[i])
{
f=0;
break;
}
}
if(!f) cout<<"NO";
else
{
cout<<"YES"<<endl;
while(!q.empty ())
{
cout<<q.front()<<endl;
q.pop();
}
}
}
return 0;
}
原文地址：https://www.cnblogs.com/caiyishuai/p/9085779.html