CF 574E(Bear and Drawing

 

E. Bear and Drawing time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Limak is a little bear who learns to draw. People usually start with houses, fences and flowers but why would bears do it? Limak lives in the forest and he decides to draw a tree.

Recall that tree is a connected graph consisting of n vertices and n - 1 edges.

Limak chose a tree with n vertices. He has infinite strip of paper with two parallel rows of dots. Little bear wants to assign vertices of a tree to some n distinct dots on a paper so that edges would intersect only at their endpoints — drawn tree must be planar. Below you can see one of correct drawings for the first sample test.

Is it possible for Limak to draw chosen tree?

Input

The first line contains single integer n (1 ≤ n ≤ 105).

Next n - 1 lines contain description of a tree. i-th of them contains two space-separated integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) denoting an edge between vertices ai and bi. It's guaranteed that given description forms a tree.

Output

Print Yes (without the quotes) if Limak can draw chosen tree. Otherwise, print No (without the quotes).

Sample test(s) input

8
1 2
1 3
1 6
6 4
6 7
6 5
7 8

output

Yes

input

13
1 2
1 3
1 4
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13

output

No

 

 

這題要構造解

構造的解是這樣的

 

 

首先我們把葉子節點以及相連的鏈標記is_lef

然後對每個點統計分出的鏈的次數(分出一顆大子樹不計)

然後除鏈和Y形，點最多分出2個大子樹，否則無解

 

 

 

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n;
int edge[MAXN],pre[MAXN],next[MAXN],siz=1;
void addedge(int u,int v) {
	edge[++siz]=v;
	next[siz]=pre[u];
	pre[u]=siz;
}
void addedge2(int u,int v){addedge(u,v),addedge(v,u);}
int degree[MAXN]={0};

bool is_lef[MAXN]={0};
int legs[MAXN]={0};
void dfs(int x,int f)
{
	is_lef[x]=1;
	Forp(x) {
		int v=edge[p];
		if (v==f) continue;
		if (degree[v]<=2) dfs(v,x); //鏈 
	}
}



int main()
{
//	freopen(E.in,r,stdin);
//	freopen(.out,w,stdout);
	
	MEM(edge) MEM(pre) MEM(next)
	cin>>n;
	For(i,n-1) {
		int u,v;
		scanf(%d%d,&u,&v);
		addedge2(u,v);
		degree[u]++;degree[v]++;
	}
	
	//找鏈 
	For(i,n) {
		if (degree[i]==1) dfs(i,0);		
	}
	
	
	//找Y型 
	For(i,n) {
		Forp(i) {
			int v=edge[p];
			if (is_lef[v]) ++legs[i]; //一個點連出的鏈數 
		}
	}
	
	
	bool flag=1;
	
	
	For(i,n) {
		if (is_lef[i]) continue;
		int cnt=0;
		
		Forp(i) {
			int v=edge[p];
			if (!is_lef[v]&°ree[v]-min(legs[v],2)>=2)  //這個鄰居不是鏈也不是Y 
				++cnt;
		} 
		
		
		if (cnt>=3) { //只能向兩邊 
			flag=0;break;
		}
		
	}
	
	
	
	if (flag) puts(Yes);
	else puts(No);
	
	
	return 0;
}