Lomsat gelral CodeForces - 600E（樹上啓發式合併）

Lomsat gelral CodeForces - 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
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

#include<bits/stdc++.h>
using namespace std;
const long long maxn = 1e5+9;
long long cnt[maxn],ans[maxn],val[maxn],sum,son[maxn],Size[maxn],flag[maxn],max_cnt;
vector<long long>edge[maxn];
void read(long long &x)
{
    x = 0;
    long long f = 1;
    char ch = getchar();
    while(!isdigit(ch))
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(isdigit(ch))
    {
        x = (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }
    x = x*f;
}

void  con_son(long long x,long long fa)
{
    long long max_num = -1;
    Size[x]=1;
    for(long long i = 0; i<edge[x].size(); i++)
    {
        long long to=edge[x][i];
        if(to==fa)
            continue;
        con_son(to,x);
        if(Size[to]>max_num)
        {
            max_num = Size[to];
            son[x] = to;
        }
        Size[x]+=Size[to];
    }
}

void conclute(long long x,long long fa,long long add)
{
    cnt[val[x]]+=add;
    if(add==1&&cnt[val[x]]>=max_cnt)
    {

        if(cnt[val[x]]>max_cnt)
            sum = val[x];
        else
            sum += val[x];
        max_cnt = cnt[val[x]];
    }
    for(long long i=0; i<edge[x].size(); i++)
    {
        long long to=edge[x][i];
        if(to==fa||flag[to])
            continue;
        conclute(to,x,add);
    }
}
void dfs(long long x,long long fa,long long save)
{
    for(long long i = 0; i<edge[x].size(); i++)
    {
        long long to=edge[x][i];
        if(to==fa||to==son[x])
            continue;
        dfs(to,x,0);
    }
    if(son[x])
    {
        dfs(son[x],x,1);
        flag[son[x]]=1;
    }
    conclute(x,fa,1);
    ans[x]=sum;
    if(son[x])
        flag[son[x]]=0;
    if(save==0)
    {
        conclute(x,fa,-1);
        sum=max_cnt=0;
    }
}
int main()
{
    long long i,j,m,n,a,b;
    ios::sync_with_stdio(false);
    read(n);
    for(i = 1; i<=n; i++)
    {
        read(val[i]);
    }
    for(i =1; i<=n-1; i++)
    {
        read(a);
        read(b);
        edge[a].push_back(b);
        edge[b].push_back(a);
    }
    con_son(1,0);
    dfs(1,0,1);
    for(i = 1; i<=n; i++)
    {
        printf("%lld%c",ans[i],i==n?'\n':' ');
    }

    return 0;
}