Educational Codeforces Round 2 E Lomsat gelral（启发式合并）

思路：维护每个子树颜色最多的数量以及每个颜色拥有的数量，然后启发式合并一波

#include<bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 1e5+7;
vector<int>e[maxn];
int n,c[maxn],num[maxn];   //
LL ans[maxn];
map<int,int>cnt[maxn];   //cnt[u][i] 以u的子树有多少种颜色i

void dfs(int u,int fa)
{
	num[u]=1;
	ans[u]=c[u];
    cnt[u][c[u]]=1;
	for(int i = 0;i<e[u].size();i++)
	{
		int v= e[u][i];
		if(v==fa)continue;
		dfs(v,u);
		if(cnt[u].size()<cnt[v].size())
		{
			swap(cnt[u],cnt[v]);
			swap(num[u],num[v]);
			ans[u]=ans[v];
		}
		for(map<int,int>::iterator it = cnt[v].begin();it!=cnt[v].end();it++)
		{
			 cnt[u][it->first]+=it->second;
             if(cnt[u][it->first] > num[u])
			 {
				 ans[u]=it->first;
				 num[u]=cnt[u][it->first];
			 }
			 else if (cnt[u][it->first] == num[u])
			 {
				 ans[u]+=it->first;
			 }
		}
	}
}
int main()
{
    scanf("%d",&n);
	for(int i = 1;i<=n;i++)
		scanf("%d",&c[i]);
	for(int i = 1;i<=n-1;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1,-1);

	for(int i = 1;i<=n;i++)
		printf("%lld%c",ans[i],i==n?'\n':' ');
}

E. Lomsat gelral

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.

Let's call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it's possible that two or more colours will be dominating in the subtree of some vertex.

The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.

For each vertex v find the sum of all dominating colours in the subtree of vertex v.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of vertices in the tree.

The second line contains n integers ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the edge of the tree. The first vertex is the root of the tree.

Output

Print n integers — the sums of dominating colours for each vertex.

Examples

input

4
1 2 3 4
1 2
2 3
2 4

output

10 9 3 4

input

15
1 2 3 1 2 3 3 1 1 3 2 2 1 2 3
1 2
1 3
1 4
1 14
1 15
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13

output

6 5 4 3 2 3 3 1 1 3 2 2 1 2 3