Submission #845893

#	Time	Username	Problem	Language	Result	Execution time	Memory
845893		sebinkim	Beech Tree (IOI23_beechtree)	C++17	0 / 100	5 ms	25692 KiB

beechtree

#include <bits/stdc++.h>
#include "beechtree.h"

using namespace std;
using pii = pair<int, int>;
using piii = tuple<int, int, int>;

vector<int> T[202020], C;
priority_queue<piii> Q;
int S[202020], I[202020];
set<pii> P1[202020];
set<piii> P2[202020];
int n, m;

bool dfs(int u, vector<int> &A) {
    bool f = 1;

    P1[u].emplace(I[u], u);
    for (int &v: T[u]) {
        auto it = P2[u].emplace(C[v], I[v], u).first;
		if (it != P2[u].begin() && get<0>(*prev(it)) == C[v]) f = 0;
		if (next(it) != P2[u].end() && get<0>(*next(it)) == C[v]) f = 0;
    }

	for (int &v: T[u]) {
		f &= dfs(v, A);

        if (P1[u].size() < P1[v].size()) {
            swap(P1[u], P1[v]); swap(P2[u], P2[v]);
        }
        for (auto [i, w]: P1[v]) {
            auto it1 = P1[u].emplace(i, w).first;
			for (int &x: T[w]) {
				auto it2 = P2[u].emplace(C[x], I[x], x).first;
				if (next(it2) != P2[u].end() && get<0>(*next(it2)) == C[x]) {
					if (next(it1) == P1[u].end() || next(it1)->second != get<2>(*next(it2))) f = 0;
				}
				if (it1 != P1[u].begin()) {
					if (it2 == P2[u].begin() || get<0>(*prev(it2)) != C[x] \
						|| get<2>(*prev(it2)) != prev(it1)->second) f = 0;
				}
			}
        }
	}

    return A[u] = f;
}

vector<int> beechtree(int n, int m, vector<int> P, vector<int> C) {
	::C = C; ::n = n; ::m = m;

	int i;

	for (i = 1; i < n; i++) {
		T[P[i]].push_back(i);
	}
    fill(S, S + n, 1);
    for (i = n - 1; i >= 1; i--) {
        S[P[i]] += S[i];
    }

	Q.emplace(S[0], 0, 0);
	for (i = 0; !Q.empty(); i++) {
		auto [_, j, v] = Q.top();
        I[v] = i; Q.pop();
		for (int &w: T[v]) {
			Q.emplace(S[w], i, w);
		}
	}

	vector<int> A(n);

	dfs(0, A);

	return A;
}

Compilation message (stderr)

beechtree.cpp: In function 'bool dfs(int, std::vector<int>&)':
beechtree.cpp:46:17: warning: suggest parentheses around assignment used as truth value [-Wparentheses]
   46 |     return A[u] = f;

• Subtask 0: 0 / 0
• Subtask 1: 0 / 9
• Subtask 2: 0 / 5
• Subtask 3: 0 / 9
• Subtask 4: 0 / 8
• Subtask 5: 0 / 14
• Subtask 6: 0 / 14
• Subtask 7: 0 / 12
• Subtask 8: 0 / 17
• Subtask 9: 0 / 12