题目链接
题解
容易想到就是线段树合并，维护每个权值区间出现的最大值以及最大值位置之和即可
对于每个节点合并一下两个子节点的信息
要注意叶子节点信息的合并和非叶节点信息的合并是不一样的
由于合并不比逐个插入复杂度高，所以应是\(O(nlogn)\)的
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 8000005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
LL sum[maxm],ans[maxn];
int ls[maxm],rs[maxm],mx[maxm],cnt;
int n,c[maxn],fa[maxn],rt[maxn];
int h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn << 1];
inline void build(int u,int v){
ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
inline void upd(int u){
sum[u] = 0;
if (mx[ls[u]] >= mx[rs[u]]){
mx[u] = mx[ls[u]];
sum[u] += sum[ls[u]];
}
if (mx[rs[u]] >= mx[ls[u]]){
mx[u] = mx[rs[u]];
sum[u] += sum[rs[u]];
}
}
void modify(int& u,int pre,int l,int r,int pos){
u = ++cnt; ls[u] = ls[pre]; rs[u] = rs[pre];
if (l == r){mx[u] = mx[pre] + 1; sum[u] = l; return;}
int mid = l + r >> 1;
if (mid >= pos) modify(ls[u],ls[pre],l,mid,pos);
else modify(rs[u],rs[pre],mid + 1,r,pos);
upd(u);
}
int merge(int u,int v,int l,int r){
if (!u) return v;
if (!v) return u;
int t = ++cnt,mid = l + r >> 1;
if (l == r){
mx[t] = mx[u] + mx[v];
sum[t] = l;
return t;
}
ls[t] = merge(ls[u],ls[v],l,mid);
rs[t] = merge(rs[u],rs[v],mid + 1,r);
upd(t);
return t;
}
void dfs(int u){
modify(rt[u],rt[u],1,n,c[u]);
Redge(u) if ((to = ed[k].to) != fa[u]){
fa[to] = u; dfs(to);
rt[u] = merge(rt[u],rt[to],1,n);
}
ans[u] = sum[rt[u]];
}
int main(){
n = read();
REP(i,n) c[i] = read();
for (int i = 1; i < n; i++) build(read(),read());
dfs(1);
REP(i,n) printf("%lld ",ans[i]);
return 0;
}