题目链接
题意：
树上每个结点有一个颜色，统计树上每个结点的子树上出现次数最多的颜色和。

思路：
树上启发式合并，将原本的\(O(n^2)\)复杂度变为\(O(n*logn)\)，OIwiki的复杂度证明。

code:

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <deque>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
// #include <unordered_map>

#define fi first
#define se second
#define pb push_back
#define endl "\n"
#define debug(x) cout << #x << ":" << x << endl;
#define bug cout << "********" << endl;
#define all(x) x.begin(), x.end()
#define lowbit(x) x & -x
#define fin(x) freopen(x, "r", stdin)
#define fout(x) freopen(x, "w", stdout)
#define ull unsigned long long
#define ll long long 

const double eps = 1e-15;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const int mod =  1e9 + 7;
const int maxn = 2e5 + 10;

using namespace std;

int L[maxn], R[maxn];
int col[maxn], dfn, num[maxn], cnt[maxn];
int sz[maxn], n, bson[maxn], sum[maxn], maxx;
ll ans[maxn], summ[maxn];
vector<int> v[maxn];

void add(int u){
    int &a = cnt[col[u]];
    sum[a] --, sum[a + 1] ++;
    summ[a] -= col[u], summ[a + 1] += col[u];
    
    cnt[col[u]] ++;
    maxx = max(maxx, cnt[col[u]]);
}

void del(int u){
    int &a = cnt[col[u]];
    sum[a] --, sum[a - 1] ++;
    summ[a] -= col[u], summ[a - 1] += col[u];

    if(cnt[col[u]] == maxx && !sum[a])maxx --;
    cnt[col[u]] --;
}

void dfs1(int u, int fa){
    sz[u] = 1;
    L[u] = ++ dfn;
    num[dfn] = u;
    for(int to : v[u]){
        if(to != fa){
            dfs1(to, u), sz[u] += sz[to];
            if(!bson[u] || sz[bson[u]] < sz[to])bson[u] = to;
        }
    }
    R[u] = dfn;
}

void dfs2(int u, int fa, bool flag){
    for(int to : v[u]){
        if(to != bson[u] && to != fa)dfs2(to, u, false);
    }
    if(bson[u])dfs2(bson[u], u, true);
    for(int to : v[u]){
        if(to != bson[u] && to != fa){
            for(int i = L[to]; i <= R[to]; i ++)add(num[i]);
        }
    }

    add(u);
    ans[u] = summ[maxx];
    if(!flag){
        for(int i = L[u]; i <= R[u]; i ++)del(num[i]);
    }
}

int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++)scanf("%d", &col[i]);
    for(int i = 1, a, b; i < n; i ++){
        scanf("%d%d", &a, &b);
        v[a].pb(b), v[b].pb(a);
    }
    dfs1(1, 0);
    dfs2(1, 0, false);
    for(int i = 1; i <= n; i ++)printf("%lld ", ans[i]);
    return 0;
}