codeforces 600E . Lomsat gelral (线段树合并）

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

Copy

41 2 3 41 22 32 4

output

Copy

10 9 3 4

input

Copy

151 2 3 1 2 3 3 1 1 3 2 2 1 2 31 21 31 41 141 152 52 62 73 83 93 104 114 124 13

output

Copy

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

思路：

对每个点建线段树在线段树上当前点的权值下标+1，每个点与父节点进行线段树合并，这样对树dfs一遍就可以求出所有的值了。

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mid ll m = (l + r) >> 1
const ll M = 1e5 + 10;

struct node{
    ll ans,sum;
}tr[M*40];

struct node1{
    ll to,next;
}e[M<<1];
ll cnt,head[M],idx,ls[M*40],rs[M*40],root[M],a[M],ans[M];

void add(ll u,ll v){
    e[++cnt].to = v;e[cnt].next = head[u];head[u] = cnt;
}

void pushup(ll rt){
    if(tr[ls[rt]].sum > tr[rs[rt]].sum){
        tr[rt].sum = tr[ls[rt]].sum;
        tr[rt].ans = tr[ls[rt]].ans;
    }
    else if(tr[ls[rt]].sum == tr[rs[rt]].sum){
        tr[rt].sum = tr[ls[rt]].sum;
        tr[rt].ans = tr[rs[rt]].ans + tr[ls[rt]].ans;
    }
    else{
        tr[rt].sum = tr[rs[rt]].sum;
        tr[rt].ans = tr[rs[rt]].ans;
    }
}

void update(ll p,ll c,ll l,ll r,ll &rt){
    if(!rt) rt = ++idx;
    if(l == r){
        tr[rt].sum += c;
        tr[rt].ans = l;
        return ;
    }
    mid;
    if(p <= m) update(p,c,l,m,ls[rt]);
    else update(p,c,m+1,r,rs[rt]);
    pushup(rt);
}

ll Merge(ll x,ll y,ll l,ll r){
    if(!x) return y;
    if(!y) return x;
    if(l == r){
        tr[x].sum += tr[y].sum;
        tr[x].ans = l;
        return x;
    }
    mid;
    ls[x] = Merge(ls[x],ls[y],l,m);
    rs[x] = Merge(rs[x],rs[y],m+1,r);
    pushup(x);
    return x;
}

void dfs(ll u,ll fa){
    for(ll i = head[u];i;i=e[i].next){
        ll v = e[i].to;
        if(v == fa) continue;
        dfs(v,u);
        Merge(root[u],root[v],1,M);
    }
    update(a[u],1,1,M,root[u]);
    ans[u] = tr[root[u]].ans;
}

int main()
{
    ll n,u,v;
    scanf("%lld",&n);
    for(ll i = 1;i <= n;i ++){
        scanf("%lld",&a[i]);
        root[i] = i;
        idx++;
    }
    for(ll i = 1;i < n;i ++){
        scanf("%lld%lld",&u,&v);
        add(u,v); add(v,u);
    }
    dfs(1,0);
    for(ll i = 1;i <= n;i ++){
        printf("%lld ",ans[i]);
    }
    return 0;

}

原文地址：https://www.cnblogs.com/kls123/p/9972034.html