codeforces 913B Christmas Spruce-CFANZ编程社区

http://www.elijahqi.win/2018/01/09/codeforces-913b-christmas-spruce/
B. Christmas Spruce
time limit per test

1 second
memory limit per test

256 megabytes
input

standard input
output

standard output

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

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:

给定一棵有根树验证每个非叶子节点的叶子结点数是否大于3 怎么搞模拟

我每次把这个边都建出来如果h[i]==0说明我是叶子节点我需要给我的父亲+1

最后都做完了循环统计一下非叶子节点的值是否都>=3即可

#include<cstdio>
#define N 1100
int h[N],par[N],n,cnt[N];
struct node{
    int y,next;
}data[N];
int main(){
    freopen("cf.in","r",stdin);
    scanf("%d",&n);int num=0;
    for (int i=2;i<=n;++i){
        int x;scanf("%d",&x);
        data[++num].y=i;data[num].next=h[x];h[x]=num;par[i]=x;
    }
    for (int i=2;i<=n;++i){
        if (!h[i]) ++cnt[par[i]];
    }bool flag=0;
    for (int i=1;i<=n;++i){
        if (h[i]&&cnt[i]<3) {flag=1;break;}
    }if(!flag) puts("Yes");else puts("No");
    return 0;
}