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
81 21 31 66 46 76 57 8
output
Yes
input
131 21 31 42 52 62 73 83 93 104 114 124 13
output
No 我们想如果一个点 连接着 大于等于两颗树是多叉树的话，那么我们就可以认为这个是无解的（除非其中一颗树是直接连接在该结点上的）。
#include<cstdio>
#include <iostream>
#include <string.h>
#include <vector>
using namespace std;
const int maxn = 100000 + 5;
vector<int> G[maxn];
int leg[maxn];
bool del[maxn];
void dfs(int cur, int per=-1)
{
if(G[cur].size()<=2)
{
del[cur]=true;
int siz=G[cur].size();
for(int i=0; i<siz; i++)
{
int b=G[cur][i];
if(b == per) continue;
dfs(b,cur);
}
}
}
int main() {
int n;
while(scanf("%d",&n)==1)
{
memset(del,false,sizeof(del));
memset(leg,0,sizeof(leg));
for(int i=1; i<=n; i++)G[i].clear();
for(int i=1; i<n; i++)
{
int a,b;
scanf("%d%d",&a,&b);
G[a].push_back(b);
G[b].push_back(a);
}
for(int i=1; i<=n; i++)
if(G[i].size() ==1 ){
dfs(i);
}
for(int a=1; a<=n; a++)
{
int siz=G[a].size();
for(int j = 0; j<siz; j++)
{
int b=G[a][j];
if(del[b])
{
leg[a]=min(leg[a]+1,2);
}
}
}
bool falg=true;
for(int a=1; a<=n; a++)
if(del[a]==false){
int cnt=0;
for(int j=0; j<G[a].size(); j++)
{
int b=G[a][j];
if(del[b]==false&&G[b].size()-leg[b]>1) cnt++;
}
if(cnt>2){
falg=false; break;
}
}
if(falg)puts("YES");
else puts("NO");
}
return 0;
}