Hello 2018 B. 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.
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.
Print "Yes" if the tree is a spruce and "No" otherwise.
4 1 1 1
Yes
7 1 1 1 2 2 2
No
8 1 1 1 1 3 3 3
Yes
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:
要求所有的点满足其中一个要求即可。①没有子节点 ②有至少三个没有子节点的子节点。
用vector来存图，之后利用size来判断即可
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include<queue>
#include<map>
using namespace std;
vector<int>mp[1100];
int dfs(int num)
{
if(mp[num].size()==0)
{
return 1;
}
int ans=0;
for(int i=0;i<mp[num].size();i++)
{
if(mp[mp[num][i]].size()==0)
{
ans++;
}
}
if(ans>=3)
return 1;
return 0;
}
int main()
{
int n;
cin>>n;
for(int i=2;i<=n;i++)
{
int u;
cin>>u;
mp[u].push_back(i);
}
int flag=0;
for(int i=1;i<=n;i++)
{
if(dfs(i)==0)
{
flag=1;
break;
}
}
if(flag)
{
cout<<"No"<<endl;
}
else
{
cout<<"Yes"<<endl;
}
}