codeforces 913 B. Christmas Spruce 【思维】

B. Christmas Spruce

time limit per test1 second

memory limit per test256 megabytes

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： 给你一个有根树，规定每个非叶子节点的孩子大于等于3，是否符合

分析： 直接用vector存下，然后利用std::vector.pop_back()，a[i].size()代表i节点的孩子节点个数可以清除最后一个元素，但这里注意我们得倒着枚举，因为这里是清除最后一个元素。

参考代码

#include<bits/stdc++.h>

using namespace std;

#define ll long long

const int N = 1e5 + 10;
vector<int> a
;

int res
;

int main(){
ios_base::sync_with_stdio(0);
int n;cin>>n;
for(int i = 2;i <= n;i++) {
int x;cin>>x;
a[x].push_back(i);
}
bool flg = false;

for(int i = 0;i < 1010;i++) {
if(a[i].size()) {
for(int j = a[i].size() - 1;j >= 0;j--) {
if(a[a[i][j]].size()) {
a[i].pop_back();
}
}
if(a[i].size() < 3) flg = true;
}
}

if(flg) cout<<"No"<<endl;
else cout<<"Yes"<<endl;

return 0;
}

如有错误或遗漏，请私聊下UP，thx