Codeforces Round #475 (Div. 2) D. Destruction of a Tree

Posted 2020-10-31 visualVK

 1 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).
 2 
 3 A vertex can be destroyed if this vertex has even degree. If you destroy a vertex, all edges connected to it are also deleted.
 4 
 5 Destroy all vertices in the given tree or determine that it is impossible.
 6 
 7 Input
 8 The first line contains integer n (1?≤?n?≤?2·105) — number of vertices in a tree.
 9 
10 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.
11 
12 Output
13 If it‘s possible to destroy all vertices, print "YES" (without quotes), otherwise print "NO" (without quotes).
14 
15 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.
16 
17 Examples
18 inputCopy
19 5
20 0 1 2 1 2
21 outputCopy
22 YES
23 1
24 2
25 3
26 5
27 4
28 inputCopy
29 4
30 0 1 2 3
31 outputCopy
32 NO

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题目大意：有n个顶点，输入n个数字，记作p[i]，如果p[i]！=0，则表示有一条i到p[i]的边，要求你输出删除顶点的顺序，删除的条件有：1。这个点必须有偶数条边才能被删除  2。跟删除点连接的边都要去掉

分析：用一个in[maxn]的数组来记录连接u点的边数是偶数还是奇数，然后用dfs来跑遍所有的路，求出所有顶点是奇数还是偶数，最后递归打印即可

  1 #define debug
  2 #include<stdio.h>
  3 #include<math.h>
  4 #include<cmath>
  5 #include<queue>
  6 #include<stack>
  7 #include<string>
  8 #include<cstring>
  9 #include<string.h>
 10 #include<algorithm>
 11 #include<iostream>
 12 #include<vector>
 13 #include<functional>
 14 #include<iomanip>
 15 #include<map>
 16 #include<set>
 17 #define pb push_back
 18 #define dbg(x) cout<<#x<<" = "<<(x)<<endl;
 19 using namespace std;
 20 typedef long long ll;
 21 typedef pair<int,int> pii;
 22 typedef pair<ll,ll>PLL;
 23 typedef pair<int,ll>Pil;
 24 const ll INF = 0x3f3f3f3f;
 25 const double inf=1e8+100;
 26 const double eps=1e-8;
 27 const int maxn =1e6;
 28 const int N = 510;
 29 const ll mod=1e9+7;
 30 //------
 31 //define
 32 bool ve[maxn];
 33 vector<int>G[maxn];
 34 int in[maxn];
 35 //dfs
 36 void dfs(int u) {
 37     for(int i=0; i<G[u].size(); i++) {
 38         int tmp=G[u][i];
 39         if(tmp==u)continue;
 40         dfs(tmp);
 41         in[u]^=in[tmp];
 42     }
 43     in[u]^=1;
 44 }
 45 void print(int u) {
 46     for(int i=0; i<G[u].size(); i++) {
 47         int tmp=G[u][i];
 48         if(tmp==u)continue;
 49         if(!in[tmp]) {
 50             print(tmp);
 51         }
 52     }
 53     cout<<u<<endl;
 54     for(int i=0; i<G[u].size(); i++) {
 55         int tmp=G[u][i];
 56         if(tmp==u)continue;
 57         if(in[tmp]) {
 58             print(tmp);
 59         }
 60     }
 61 }
 62 //solve
 63 void solve() {
 64     int n,s;
 65     cin>>n;
 66     for(int i=0; i<n; i++) {
 67         int u;
 68         cin>>u;
 69         if(u) {
 70             G[u].push_back(i+1);
 71         } else {
 72             s=i+1;
 73         }
 74     }
 75     dfs(s);
 76     if(in[s]) {
 77         cout<<"YES"<<endl;
 78         print(s);
 79     }else{
 80         cout<<"NO"<<endl;
 81     }
 82 }
 83 //main
 84 int main() {
 85     ios_base::sync_with_stdio(false);
 86 #ifdef debug
 87     freopen("in.txt", "r", stdin);
 88 //    freopen("out.txt","w",stdout);
 89 #endif
 90     cin.tie(0);
 91     cout.tie(0);
 92     solve();
 93     /*
 94         #ifdef debug
 95             fclose(stdin);
 96             fclose(stdout);
 97             system("out.txt");
 98         #endif
 99     */
100     return 0;
101 }

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