[树的度数] Christmas Spruce

Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called a child of vertex v and vertex v is called a parent of vertex u if there exists a directed edge from v to u. A vertex is called a leaf if it doesn't have children and has a parent.

Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf children. You are given a rooted tree, check whether it's a spruce.

The definition of a rooted tree can be found here.


Input

The first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000). Each of the next n - 1 lines contains one integer pi (1 ≤ i ≤ n - 1) — the index of the parent of the i + 1-th vertex (1 ≤ pi ≤ i).

Vertex 1 is the root. It's guaranteed that the root has at least 2 children.

Output

Print "Yes" if the tree is a spruce and "No" otherwise.

Examples
Input 
4
1
1
1
Output
Yes
Input
7
1
1
1
2
2
2
Output
No
Input
8
1
1
1
1
3
3
3
Output
Yes
Note

The first example:

The second example:

It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:

题意:问一棵树的所有非叶子结点是否至少有三个叶子
思路:BFS搜索每个非叶子结点的叶子结点的个数,碰到少于3的输出No,否则若都大于等于3个就输出Yes

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<queue>
 5 using namespace std;
 6 typedef long long ll;
 7 const int amn=1e5+5;
 8 int n,ans=0,m[amn],deep[amn],idx[amn],cnt;
 9 vector<int> eg[amn];
10 queue<int> q;
11 int bfs(int rt){
12     while(q.size())q.pop();q.push(rt);
13     memset(deep,0,sizeof deep);
14     memset(idx,0,sizeof idx);
15     deep[rt]=1;
16     cnt=0;
17     while(q.size()){
18         int u=q.front();q.pop();
19         idx[u]=1;
20         cnt=0;
21         for(int i=0;i<eg[u].size();i++){
22             int v=eg[u][i];
23             if(idx[v]||v==u)continue;
24             if(!eg[v].size())cnt++; ///统计叶子结点个数
25         }
26         if(cnt>=3){
27                 for(int i=0;i<eg[u].size();i++){
28                 int v=eg[u][i];
29                 if(idx[v]||v==u)continue;
30                 if(eg[v].size())q.push(v);
31             }
32         }
33         else return 0;
34     }
35     return 1;
36 }
37 int main(){
38     scanf("%d",&n);
39     for(int i=2;i<=n;i++){
40         scanf("%d",&m[i]);
41         eg[m[i]].push_back(i);
42     }
43     ans=bfs(1);
44     if(ans)
45     printf("Yes\n");
46     else printf("No\n");
47 }
48 /**
49 题意:问一棵树的所有非叶子结点是否至少有三个叶子
50 思路:BFS搜索每个非叶子结点的叶子结点的个数,碰到少于3的输出No,否则若都大于等于3个就输出Yes
51 **/

 

posted @ 2019-11-09 19:32  Railgun000  阅读(185)  评论(0)  编辑  收藏  举报