CCPC Final 2019 K. Russian Dolls on the Christmas Tree

发表于 更新于 本文字数： 539 阅读时长 ≈ 2 分钟 
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

struct lx{
	mutable int l,r;
	bool operator <(const lx& aa)const{
		return r==aa.r?l<aa.l:r<aa.r;
	}
};

ll tt,n,son[(int)2e5+9],sz[(int)2e5+9],ans,res[(int)2e5+9],cnt[(int)2e5+9];
vector<int> g[(int)2e5+9];
set<lx> s;

void dfs(int u,int fa){
	for(auto it:g[u])if(it!=fa){
		dfs(it,u);
		sz[u]+=sz[it],son[u]=sz[son[u]]<sz[it]?it:son[u];
	}
}

void cal(int u,int fa,int bson,int sta){
	/*if(sta){
		auto it=s.lower_bound({u,u});
		if(it!=s.begin()&&it!=s.end()){
			auto pr=prev(it);
			if(pr->r==u-1&&it->l==u+1){
				int lll=pr->l,rrr=it->r;
				s.erase(pr,next(it)),s.insert({lll,rrr});
			}else if(pr->r==u-1)pr->r=u;
			else if(pr->l==u+1)pr->l=u;
			else s.insert({u,u});
		}else{
			if(it!=s.end()&&it->l==u+1)it->l=u;
			else if(it!=s.begin()){
				it=prev(it);
				if(it->r==u-1)it->r=u;
				else s.insert({u,u});
			}else s.insert({u,u});
		}
	}else{
		auto it=s.lower_bound({u,u});
		if(it->r==it->l&&it->l==u){
			s.erase(it);
		}else if(it->r==u){
			it->r=u-1;
		}else if(it->l==u){
			it->l=u+1;
		}else {
			int tp=it->r;
			it->r=u-1;
			s.insert({u+1,tp});
		}
	}*/
	if(sta){
		if(cnt[u-1]&&cnt[u+1])cnt[u]=1,ans-=1;
		else if(cnt[u-1]||cnt[u+1])cnt[u]=1;
		else cnt[u]=1,ans+=1;
	}else{
		if(cnt[u-1]&&cnt[u+1])cnt[u]=0,ans+=1;
		else if(cnt[u-1]||cnt[u+1])cnt[u]=0;
		else cnt[u]=0,ans-=1;
	}
	for(auto it:g[u])if(it!=bson&&it!=fa)cal(it,u,bson,sta);
}

void dfs2(int u,int fa,int sta){
	for(auto it:g[u])if(it!=fa&&it!=son[u])dfs2(it,u,0);
	if(son[u])dfs2(son[u],u,1);
	cal(u,fa,son[u],1);
	res[u]=ans;
	if(!sta)cal(u,fa,0,0);
}

void solve(){
	cin>>n;
	ans=cnt[0]=cnt[n+1]=cnt[n+2]=0;
	f(i,1,n)g[i].clear(),sz[i]=1,son[i]=0,cnt[i]=0;
	for(int i=1,x,y;i<n;++i){
		cin>>x>>y;
		g[x].push_back(y),g[y].push_back(x);
	}
	dfs(1,0);
	dfs2(1,0,0);
	for(int i=1;i<=n;++i)cout<<res[i]<<" \n"[i==n];
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>tt;
    f(ss,1,tt){
    	cout<<"Case #"<<ss<<": ";
    	solve();
    }
    return 0;
}