启发式合并 CodeForces - 600E

启发式合并最重要的思想就是指的是每次将小集合拷贝合并至大集合。
考虑每个元素的合并开销。对于合并次数最多的那个元素来说，它每合并一次，所在集合的规模扩大两倍，最多只会合并 logN 次，因而对于所有元素,至多插入set中NlogN次，时间复杂度就有上限 O(N∗logN∗logN)

现在主要是dsu on tree

但是dsu on tree也有一定的局限性：1.只能支持子树查询；2.不支持修改修改。

看例题来说：

D - Lomsat gelralCodeForces - 600E

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 c_i (1 ≤ c_i ≤ n), c_i — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers x_j, y_j (1 ≤ x_j, y_j ≤ 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

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

题意：给定一个  $n$

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long LL;
const int maxn=1e5+5;
int n,m;
int mx,big;
LL sum=0;
LL ans[maxn];
int col[maxn],si[maxn],hson[maxn],cnt[maxn];
vector<int>G[maxn];

void findhson(int x,int fa)//找到所有的重儿子
{
    si[x]=1;
    int len=G[x].size();
    for(int i=0;i<len;i++)
    {
        int t=G[x][i];
        if(t!=fa)
        {
            findhson(t,x);
            si[x]+=si[t];
            if(si[t]>si[hson[x]])
                hson[x]=t;
        }
    }
}
void cal(int x,int fa,int val)//计算的过程，就是不断递归求颜色的数量
{
    cnt[col[x]]+=val;
    if(cnt[col[x]]>mx)
    {
        sum=col[x];
        mx=cnt[col[x]];
    }
    else if(cnt[col[x]]==mx)
        sum+=col[x];
    int len=G[x].size();
    for(int i=0;i<len;i++)
    {
        int t=G[x][i];
        if(t!=fa && t!=big)
            cal(t,x,val);
    }
}
void dfs(int x,int fa,int flag)//flag作为标记，看是轻子树还是重子树
{
    int len=G[x].size();
    for(int i=0;i<len;i++)//先跑轻子树
    {
        int t=G[x][i];
        if(t!=fa && t!=hson[x])
            dfs(t,x,0);
    }
    if(hson[x])//再跑重子树
    {
        dfs(hson[x],x,1);
        big=hson[x];
    }
    cal(x,fa,1);
    big=0;
    ans[x]=sum;
    if(!flag)//如果是轻子树的话，消除影响
    {
        cal(x,fa,-1);
        mx=0;sum=0;
    }
}
int main()
{
    sum=0;big=0;mx=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&col[i]);
    int x,y;
    for(int i=1;i<n;i++)//双向边
    {
        scanf("%d %d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    findhson(1,0);
    dfs(1,0,1);
    for(int i=1;i<=n;i++)
    {
        if(i==1)
            printf("%lld",ans[i]);
        else
            printf(" %lld",ans[i]);
    }
    printf("\n");
    return 0;
}

$n$

启发式合并 CodeForces - 600E

D - Lomsat gelralCodeForces - 600E

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