結果

問題 No.2634 Tree Distance 3
ユーザー kaikey
提出日時 2024-02-16 22:36:06
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,082 ms / 3,000 ms
コード長 5,845 bytes
コンパイル時間 3,166 ms
コンパイル使用メモリ 216,848 KB
最終ジャッジ日時 2025-02-19 14:19:30
ジャッジサーバーID
(参考情報)
judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 69
権限があれば一括ダウンロードができます

ソースコード

diff #

#include "bits/stdc++.h"
#include <random>
#include <chrono>
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define SZ(x) ((lint)(x).size())
#define FOR(i, begin, end) for(lint i=(begin),i##_end_=(end);i<i##_end_;++i)
#define IFOR(i, begin, end) for(lint i=(end)-1,i##_begin_=(begin);i>=i##_begin_;--i)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define endk '\n'
using namespace std; typedef unsigned long long _ulong; typedef long long int lint; typedef long double ld; typedef pair<lint, lint> plint; typedef pair<ld, ld> pld;
struct fast_ios { fast_ios() { cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(30); }; } fast_ios_;
template<class T> auto add = [](T a, T b) -> T { return a + b; };
template<class T> auto mul = [](T a, T b) -> T { return a * b; };
template<class T> auto f_max = [](T a, T b) -> T { return max(a, b); };
template<class T> auto f_min = [](T a, T b) -> T { return min(a, b); };
template<class T> using V = vector<T>;
using Vl = V<lint>; using VVl = V<Vl>; using VVVl = V<V<Vl>>;
template< typename T > ostream& operator<<(ostream& os, const vector< T >& v) {
    for (int i = 0; i < (int)v.size(); i++) os << v[i] << (i + 1 != v.size() ? " " : "");
    return os;
}
template< typename T >istream& operator>>(istream& is, vector< T >& v) {
    for (T& in : v) is >> in;
    return is;
}
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template <class T>
T div_floor(T a, T b) {
    if (b < 0) a *= -1, b *= -1;
    return a >= 0 ? a / b : (a + 1) / b - 1;
}
template <class T>
T div_ceil(T a, T b) {
    if (b < 0) a *= -1, b *= -1;
    return a > 0 ? (a - 1) / b + 1 : a / b;
}
template <class F> struct rec {
    F f;
    rec(F&& f_) : f(std::forward<F>(f_)) {}
    template <class... Args> auto operator()(Args &&... args) const {
        return f(*this, std::forward<Args>(args)...);
    }
};
lint gcd(lint a, lint b) { if (b == 0) return a; else return gcd(b, a % b); }
lint digit(lint a) { return (lint)log10(a); }
lint e_dist(plint a, plint b) { return abs(a.first - b.first) * abs(a.first - b.first) + abs(a.second - b.second) * abs(a.second - b.second); }
lint m_dist(plint a, plint b) { return abs(a.first - b.first) + abs(a.second - b.second); }
bool check_overflow(lint a, lint b, lint limit) { if (b == 0) return false; return a > limit / b; } // a * b > c => true
void Worshall_Floyd(VVl& g) { REP(k, SZ(g)) REP(i, SZ(g)) REP(j, SZ(g)) chmin(g[i][j], g[i][k] + g[k][j]); }
const lint MOD1000000007 = 1000000007, MOD998244353 = 998244353, INF = 2e18;
lint dx[8] = { 0, 1, 0, -1, 1, -1, 1, -1 }, dy[8] = { 1, 0, -1, 0, -1, -1, 1, 1 };
bool YN(bool flag) { cout << (flag ? "YES" : "NO") << endk; return flag; } bool yn(bool flag) { cout << (flag ? "Yes" : "No") << endk; return flag; }
struct Edge {
    lint from, to;
    lint cost;
    Edge() {

    }
    Edge(lint u, lint v, lint c) {
        cost = c;
        from = u;
        to = v;
    }
    bool operator<(const Edge& e) const {
        return cost < e.cost;
    }
};
struct WeightedEdge {
    lint to;
    lint cost;
    WeightedEdge(lint v, lint c) {
        to = v;
        cost = c;
    }
    bool operator<(const WeightedEdge& e) const {
        return cost < e.cost;
    }
};
using WeightedGraph = V<V<WeightedEdge>>;
typedef pair<lint, plint> tlint;
typedef pair<ld, ld> pld;
typedef pair<plint, plint> qlint;
typedef pair<Edge, lint> pEd;
typedef pair<plint, V<plint>> vVl;
typedef pair<string, string> pstr;
typedef pair<ld, lint> pint;

template< typename G >
struct DoublingLowestCommonAncestor {
	int LOG = 2;
	vector< int > dep;
	const G& g;
	vector< vector< int > > table;

	DoublingLowestCommonAncestor(const G& g) : g(g), dep(g.size()) {
		int curr = 1;
		while (curr < g.size()) {
			curr *= 2;
			LOG++;
		}
		table.assign(LOG, vector< int >(g.size(), -1));
	}

	void dfs(int idx, int par, int d) {
		table[0][idx] = par;
		dep[idx] = d;
		for (auto& to : g[idx]) {
			if (to != par) dfs(to, idx, d + 1);
		}
	}

	void build() {
		dfs(0, -1, 0);
		for (int k = 0; k + 1 < LOG; k++) {
			for (int i = 0; i < table[k].size(); i++) {
				if (table[k][i] == -1) table[k + 1][i] = -1;
				else table[k + 1][i] = table[k][table[k][i]];
			}
		}
	}

	int query(int u, int v) {
		if (dep[u] > dep[v]) swap(u, v);
		for (int i = LOG - 1; i >= 0; i--) {
			if (((dep[v] - dep[u]) >> i) & 1) v = table[i][v];
		}
		if (u == v) return u;
		for (int i = LOG - 1; i >= 0; i--) {
			if (table[i][u] != table[i][v]) {
				u = table[i][u];
				v = table[i][v];
			}
		}
		return table[0][u];
	}
};

void solve() {
	lint N;
	cin >> N;
	Vl arr(N);
	cin >> arr;
	map<lint, Vl> fx;
	REP(i, N) {
		fx[arr[i]].push_back(i);
	}

	VVl to(N);
	REP(i, N - 1) {
		lint u, v;
		cin >> u >> v; u--; v--;
		to[u].push_back(v);
		to[v].push_back(u);
	}

	DoublingLowestCommonAncestor lca(to);
	lca.build();

	Vl ans(N);
	plint edge = { -1, -1 };
	auto itr = fx.rbegin();

	auto dist = [&](lint v, lint u) {
		return lca.dep[u] + lca.dep[v] - lca.dep[lca.query(u, v)] * 2;
	};

	while (itr != fx.rend()) {
		for (auto v : (*itr).second) {
			if (edge.first == -1) edge.first = v;
			else if (edge.second == -1) edge.second = v;
			else {
				Vl _arr = { edge.first, edge.second, v };
				lint maxv = 0;
				REP(i, 3) {
					FOR(j, i + 1, 3) {
						if (chmax(maxv, (lint)dist(_arr[i], _arr[j]))) {
							edge = { _arr[i], _arr[j] };
						}
					}
				}
			}
		}
		for (auto v : (*itr).second) {
			if (edge.first != -1) chmax(ans[v], (lint)dist(v, edge.first));
			if (edge.second != -1) chmax(ans[v], (lint)dist(v, edge.second));
		}
		itr++;
	}

	cout << ans << endk;
}

int main() {
    lint T = 1;
    //cin >> T;
    while (T--) solve();
}
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