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