Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2) E DFS

連結：戳這裡

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).

Examples
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

題意：

給出一棵樹，然後問你能不能在兩行格點裡面畫出這棵樹

思路：

首先分析一下，對於一個根節點如果有三顆或者三顆以上的子樹的葉子節點>=3，肯定是不行的

還需要考慮的是，當一個根節點已經有了兩個子樹的葉子節點>=3了，現在在來一顆子樹，但是這顆子樹不是鏈的話，也不行。然後就深搜吧

對於節點大於等於3的好說，但是需要處理是不是鏈的。嗯 我還是看了看別人的處理方式。自己太弱了

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n;
struct edge{
    int v,next;
}e[500100];
int head[100100],tot=0;
void Add(int u,int v){
    e[tot].v=v;
    e[tot].next=head[u];
    head[u]=tot++;
}
int res[100100];
int deep[100100];
int vis[100100];
void DFS(int u){
    vis[u]=1;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(deep[v]<=2 && !vis[v]) DFS(v);
    }
}
int main(){
    mst(head,-1);
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        Add(u,v);
        Add(v,u);
        deep[u]++;
        deep[v]++;
    }
    for(int i=1;i<=n;i++){
        if(deep[i]==1) DFS(i);
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]){
            for(int j=head[i];j!=-1;j=e[j].next){
                int v=e[j].v;
                if(vis[v]){
                    res[i]=min(res[i]+1,2);
                }
            }
        }
    }
    for(int u=1;u<=n;u++){
        if(!vis[u]){
            int cnt=0;
            for(int i=head[u];i!=-1;i=e[i].next){
                int v=e[i].v;
                if(!vis[v] && deep[v]-res[v]>1) cnt++;
            }
            if(cnt>2) {
                cout<<"No"<<endl;
                return 0;
            }
        }
    }
    cout<<"Yes"<<endl;
    return 0;
}
/*
20
5 15
20 4
11 18
1 14
18 2
14 17
8 10
13 1
11 6
14 16
12 8
9 3
13 15
8 17
3 13
1 18
17 7
9 20
19 12

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