cp/codeforces/1003/f.cc
2025-02-09 15:59:43 -05:00

134 lines
2.4 KiB
C++

// {{{
#include <bits/stdc++.h>
// https://codeforces.com/blog/entry/96344
#pragma GCC optimize("O2,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
using namespace std;
template <typename T>
constexpr T MIN = std::numeric_limits<T>::min();
template <typename T>
constexpr T MAX = std::numeric_limits<T>::max();
template <typename T>
[[nodiscard]] static T sc(auto &&x) {
return static_cast<T>(x);
}
template <typename T>
[[nodiscard]] static T sz(auto &&x) {
return static_cast<T>(x.size());
}
template <typename... Args>
void pr(std::format_string<Args...> fmt, Args &&...args) {
std::print(fmt, std::forward<Args>(args)...);
}
template <typename... Args>
void pr(std::format_string<Args...> fmt) {
std::print(fmt);
}
template <typename... Args>
void prln(std::format_string<Args...> fmt, Args &&...args) {
std::println(fmt, std::forward<Args>(args)...);
}
template <typename... Args>
void prln(std::format_string<Args...> fmt) {
std::println(fmt);
}
void prln() {
std::println();
}
void prln(auto const &t) {
std::println("{}", t);
}
using ll = long long;
using ld = long double;
template <typename T>
using vec = std::vector<T>;
#define ff first
#define ss second
#define eb emplace_back
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
// }}}
void solve() {
int n;
cin >> n;
vec<int> a(n + 1);
for (int i = 1; i < sz<int>(a); ++i) {
cin >> a[i];
}
vec<vec<int>> tree(n + 1, vec<int>());
string ans(n, '0');
int root;
for (int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
root = u;
tree[u].pb(v);
tree[v].pb(u);
}
unordered_map<int, unordered_map<int, int>> children;
vec<int> par(n + 1, -1);
par[root] = root;
queue<int> q{{root}};
while (!q.empty()) {
int size = sz<int>(q);
while (size--) {
auto u = q.front();
q.pop();
for (auto v : tree[u]) {
if (par[v] != -1)
continue;
par[v] = u;
q.push(v);
if (++children[u][a[v]] == 2)
ans[a[v] - 1] = '1';
}
}
}
q.push(root);
for (int u = 1; u <= n; ++u) {
if (par[u] != u && a[u] == a[par[u]] ||
par[u] != root && a[u] == a[par[par[u]]])
ans[a[u] - 1] = '1';
}
prln("{}", ans);
}
// {{{
int main() {
cin.tie(nullptr)->sync_with_stdio(false);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
// }}}