题意
给定一棵根为 \(1\) 的 有根树。
每个节点有颜色，求每个节点子树内出现最多的颜色编号之和。
Sol
Dsu on tree板子题。
首先对于整棵树进行轻重链剖分，注意到一个关键性质：轻边只有 \(log\) 条。
\(n ^ 2\) 的暴力是 \(trivial\) 的，不再赘述。
注意到中间有很多节点被重复计算了多次。
考虑如何减少重复计算的次数。
不难想到保留重儿子的贡献，每次重新计算轻儿子的贡献。
结合上述性质，不难发现时间复杂度为 \(O(n log n)\)
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <queue>
#define int long long
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
const int N = 1e5 + 5, M = 2e5 + 5;
namespace G {
array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
cnt++;
nex[cnt] = fir[x];
to[cnt] = y;
fir[x] = cnt;
}
}
array <int, N> s;
array <int, N> ans;
namespace Dsu {
using G::fir; using G::nex; using G::to;
array <int, N> siz, dep, fa, son;
array <int, N> dfn, idx;
array <int, N> cur;
queue <int> q;
int tot, _ans;
void add(int x) {
cur[s[x]]++;
q.push(s[x]);
if (cur[s[x]] > tot) tot = cur[s[x]], _ans = s[x];
else if (cur[s[x]] == tot) _ans += s[x];
}
void clear() {
while (!q.empty())
cur[q.front()] = 0, q.pop();
tot = 0, _ans = 0;
}
int cnt;
void dfs1(int x) {
cnt++;
idx[cnt] = x;
dfn[x] = cnt;
siz[x] = 1;
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x]) continue;
fa[to[i]] = x;
dep[to[i]] = dep[x] + 1;
dfs1(to[i]);
siz[x] += siz[to[i]];
if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
}
}
void dfs2(int x) {
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x] || to[i] == son[x]) continue;
dfs2(to[i]), clear();
}
if (son[x]) dfs2(son[x]);
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x] || to[i] == son[x]) continue;
for (int j = dfn[to[i]]; j <= dfn[to[i]] + siz[to[i]] - 1; j++)
add(idx[j]);
}
add(x);
ans[x] += _ans;
}
}
signed main() {
int n = read();
for (int i = 1; i <= n; i++)
s[i] = read();
for (int i = 2; i <= n; i++) {
int x = read(), y = read();
G::add(x, y), G::add(y, x);
}
Dsu::dfs1(1), Dsu::dfs2(1);
for (int i = 1; i <= n; i++)
write(ans[i]), putchar(32);
puts("");
return 0;
}