codeforces600 E. Lomsat gelral（启发式合并）

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

Copy

4
1 2 3 4
1 2
2 3
2 4

output

Copy

10 9 3 4

input

Copy

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

Copy

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

题意：

给出一棵树和每个结点的颜色，求以 i 为根节点的子树中最多的颜色数量和

dsu on tree的板子题，讲解：https://www.cnblogs.com/zwfymqz/p/9683124.html

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 1e5 + 10;
int col[N], son[N], cnt[N], siz[N], Son, Mx;
ll ans[N], sum;
vector<int> mp[N];

void init() {
    sum = 0;
    memset(son, 0, sizeof(son));
    memset(cnt, 0, sizeof(cnt));
    memset(siz, 0, sizeof(siz));
}

void dfs(int u, int fa) {
    siz[u] = 1;
    int sz = mp[u].size();
    for(int i = 0; i < sz; ++i) {
        int v = mp[u][i];
        if(v == fa) continue;
        dfs(v, u);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]]) son[u] = v;
    }
}

void add(int u, int fa, int val) {
    cnt[col[u]] += val;
    if(cnt[col[u]] > Mx) {
        Mx = cnt[col[u]];
        sum = col[u];
    }
    else if(cnt[col[u]] == Mx)
        sum += 1ll * col[u];
    int sz = mp[u].size();
    for(int i = 0; i < sz; ++i) {
        int v = mp[u][i];
        if(v == fa || v == Son) continue;
        add(v, u, val);
    }
}

void dfs2(int u, int fa, int op) {
    int sz = mp[u].size();
    for(int i = 0; i < sz; ++i) {
        int v = mp[u][i];
        if(v == fa) continue;
        if(v != son[u]) dfs2(v, u, 0);
    }
    if(son[u]) {
        dfs2(son[u], u, 1);
        Son = son[u];
    }
    add(u, fa, 1);
    Son = 0;
    ans[u] = sum;
    if(!op) {
        add(u, fa, -1);
        sum = 0;
        Mx = 0;
    }
}

int main() {
    int n, u, v;
    while(~scanf("%d", &n)) {
        init();
        for(int i = 1; i <= n; ++i) mp[i].clear();
        for(int i = 1; i <= n; ++i) scanf("%d", &col[i]);
        for(int i = 1; i < n; ++i) {
            scanf("%d%d", &u, &v);
            mp[u].push_back(v);
            mp[v].push_back(u);
        }
        dfs(1, 0);
        dfs2(1, 0, 0);
        for(int i = 1; i <= n; ++i) {
            if(i > 1) printf(" ");
            printf("%lld", ans[i]);
        }
        printf("\n");
    }
    return 0;
}