「CF1109F」Dominant Indices【樹上啓發式合併】

F. Dominant Indices

time limit per test 4.5 seconds memory limit per test 512 megabytes input standard input output standard output

You are given a rooted undirected tree consisting of $n$ vertices. Vertex $1$ is the root.

Let’s denote a depth array of vertex $x$ as an infinite sequence $[d_{x,0},d_{x,1},d_{x,2},…],$ where $d_{x,i}$ is the number of vertices $y$ such that both conditions hold:

• $x$ is an ancestor of $y$;
• the simple path from $x$ to $y$ traverses exactly $i$ edges.

The dominant index of a depth array of vertex $x$ (or, shortly, the dominant index of vertex $x$) is an index $j$ such that:

• for every $k<j, d_{x,k}<d_{x,j}$;
• for every $k>j, d_{x,k}\leq d_{x,j}.$

For every vertex in the tree calculate its dominant index.

Input

The first line contains one integer $n\ (1\leq n\leq 10^6)$ — the number of vertices in a tree.

Then $n−1$ lines follow, each containing two integers $x$ and $y$ $(1\leq x,y\leq n, x!=y).$ This line denotes an edge of the tree.

It is guaranteed that these edges form a tree.

Output

Output $n$ numbers. $i-th$ number should be equal to the dominant index of vertex $i$.

Examples

input

4
1 2
2 3
3 4

output

0
0
0
0

input

4
1 2
1 3
1 4

output

1
0
0
0

input

4
1 2
2 3
2 4

output

2
1
0
0

題意

• 就是給你一顆有根樹，對於每一顆子樹，統計每一個深度下的節點數，求有最多節點的深度$h$，如果節點數有相同的選深度小的那個

題解

• 樹上啓發式合併模板題
• 對於相同節點的深度，可以直接用$10^6$個$set$去維護節點數爲$i$的所有深度，然後就能方便的去撤銷子樹的信息
• 複雜度$n(\log n)^2$，$4.5s$剛剛好

代碼

#include<bits/stdc++.h>
 
using namespace std;
const int maxn=2e6+10;

//題目數據
int t,n,u,v,q,k,a[maxn],b[maxn],ans[maxn],cnt[maxn],sum,maxx=0,he=0,cnth[maxn],pos[maxn];
vector<int> vec[maxn];

//樹剖用
int tot=0,siz[maxn],son[maxn];
int vis[maxn];

void dfs1(int cur,int fath,int he){ //dfs(root,0,1)
    siz[cur]=1;
    for(int i=0;i<vec[cur].size();i++){
        if(vec[cur][i]!=fath){
            dfs1(vec[cur][i],cur,he+1);
            siz[cur]+=siz[vec[cur][i]];
            if(siz[vec[cur][i]]>siz[son[cur]]) son[cur]=vec[cur][i];
        }
    }
}

void calc(int cur,int fa,int val,int h)
{
    if(val==1) {
        if(cnth[h]) pos[cnth[h]].erase(h);
        cnth[h]++;
        pos[cnth[h]].insert(h);
        if(cnth[h]>maxx) {maxx=cnth[h];he=(*pos[cnth[h]].begin());}
        else if(cnth[h]==maxx) he=(*pos[cnth[h]].begin());
    }else {
        pos[cnth[h]].erase(h);
        cnth[h]--;
        pos[cnth[h]].insert(h);
        if(!pos[cnth[h]+1].size()&&maxx==cnth[h]+1) {
            maxx=cnth[h];
            he=(*pos[cnth[h]].begin());
        }else if(maxx==cnth[h]) {
            he=(*pos[cnth[h]].begin());
        }
    }
    for(int i=0;i<vec[cur].size();i++){
        if(vec[cur][i]!=fa&&!vis[vec[cur][i]]){  
            calc(vec[cur][i],cur,val,h+1);
        }
    }
}

void dfs(int cur,int fa,bool keep,int h) 
{
    for(int i=0;i<vec[cur].size();i++){
        if(vec[cur][i]!=fa&&vec[cur][i]!=son[cur]){
            dfs(vec[cur][i],cur,0,h+1);
        }
    }
    if(son[cur]) dfs(son[cur],cur,1,h+1),vis[son[cur]]=1; 
    calc(cur,fa,1,h);
    ans[cur]=he-h; 
    if(son[cur]) vis[son[cur]]=0;  
    if(!keep) calc(cur,fa,-1,h);  
}

int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        scanf("%d %d",&u,&v);
        vec[u].push_back(v);
        ec[v].push_back(u);
    }
    dfs1(1,0,1);
    dfs(1,0,0,1);
    for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
}