Heuristic merging on tree

3189. Lomsat gelral

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
const double eps=1e-7;
#define x first
#define y second
#define int long long
#define lb long double
#define pb push_back
#define endl '\n'
#define all(v) (v).begin(),(v).end()
#define PII pair<int,int>
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define dwn(i,n,x) for(int i=n;i>=x;i--)
#define ll_INF 0x7f7f7f7f7f7f7f7f
#define INF 0x3f3f3f3f
#define debug(x) cerr << #x << ": " << x << endl
#define io ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
int Mod(int a,int mod){
    return (a%mod+mod)%mod;}
int lowbit(int x){
    return x&-x;}// Its lowest 1 And behind it 0 The value of composition 
int qmi(int a, int k, int p){
    int res = 1 % p;while (k){
    if (k & 1) res = Mod(res * a , p);a = Mod(a * a , p);k >>= 1;}return res;}
int inv(int a,int mod){
    return qmi(a,mod-2,mod);}
const int N=100010,M=N*2;
int n;
int h[N],e[M],ne[M],idx;
int color[N],cnt[N],sz[N],son[N];
int ans[N],sum; 
int maxv;
void add(int a,int b)
{
    
	e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
int dfs_son(int u,int father)// Heavy chain subdivision 
{
    
	sz[u]=1;
	for(int i=h[u];~i;i=ne[i])
	{
    
		int j=e[i];
		if(j==father)continue;
		sz[u]+=dfs_son(j,u);
		if(sz[j]>sz[son[u]])son[u]=j;
	}
	return sz[u];
}
void update(int u,int father,int sign,int pson)
{
    
	int c=color[u];
	cnt[c]+=sign;
	if(cnt[c]>maxv)maxv=cnt[c],sum=c;
	else if(cnt[c]==maxv)sum+=c;
	for(int i=h[u];~i;i=ne[i])
	{
    
		int j=e[i];
		if(j==father||j==pson)continue;
		update(j,u,sign,pson);
	}
}
void dfs(int u,int father,int op)// The contribution of the light side of violence statistics ,op==0 It means that the influence on this point is eliminated after recursion 
{
    
	for(int i=h[u];~i;i=ne[i])
	{
    
		int j=e[i];
		if(j==father||j==son[u])continue;
		dfs(j,u,0);
	}
	if(son[u])dfs(son[u],u,1);
	update(u,father,1,son[u]);
	ans[u]=sum;
	if(!op)update(u,father,-1,0),sum=maxv=0;
}
void solve()
{
    
	cin>>n;
	rep(i,1,n)cin>>color[i];
	memset(h,-1,sizeof h);
	rep(i,0,n-2)
	{
    
		int a,b;
		cin>>a>>b;
		add(a,b),add(b,a);
	}
	dfs_son(1,-1);
	dfs(1,-1,1);
	rep(i,1,n)cout<<ans[i]<<' ';
}
signed main()
{
    
    io;
    int _;_=1;
   // cin>>_;
    while(_--)solve();
    return 0;
}