제출 #526660

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
526660		jwvg0425	낙하산 고리들 (IOI12_rings)	C++17	37 / 100	1138 ms	107400 KiB

rings

#include <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
#include <iostream>
#include <string>
#include <bitset>
#include <map>
#include <set>
#include <tuple>
#include <string.h>
#include <math.h>
#include <random>
#include <functional>
#include <assert.h>
#include <math.h>
#define all(x) (x).begin(), (x).end()
#define xx first
#define yy second

using namespace std;

template<typename T, typename Pr = less<T>>
using pq = priority_queue<T, vector<T>, Pr>;
using i64 = long long int;
using ii = pair<int, int>;
using ii64 = pair<i64, i64>;

int n;

void Init(int n_)
{
	n = n_;
}

vector<int> edge[1000005];
int order[1000005];
int t = 0;
int par[1000005];
int backStMax;
int backEdMin;
vector<int> s;

void Link(int a, int b)
{
	edge[a].push_back(b);
	edge[b].push_back(a);
}

void dfs(int x)
{
	s.push_back(x);
	t++;
	order[x] = t;

	for (auto& e : edge[x])
	{
		if (order[e] == 0)
		{
			par[e] = x;
			dfs(e);
		}
		else if (e != par[x] && order[e] < order[x])
		{
			backStMax = max(backStMax, order[e]);
			backEdMin = min(backEdMin, order[x]);
		}
	}
}

int getDegree(int x, int off = 0)
{
	return min((int)edge[x].size() - off, 3);
}

int solve(int x)
{
	s.clear();
	backStMax = 0;
	backEdMin = 987654321;
	dfs(x);

	vector<int> cnt(6);

	for (auto& si : s)
		cnt[getDegree(si)]++;

	if (cnt[0] != 0 || (cnt[1] == 2 && cnt[3] == 0))
		return 0;

	int res = 0;

	for (auto& si : s)
	{
		cnt[getDegree(si)]--;
		for (auto& e : edge[si])
		{
			cnt[getDegree(e)]--;
			cnt[getDegree(e, 1)]++;
		}

		if (cnt[3] == 0 && order[si] >= backStMax && order[si] <= backEdMin)
			res++;

		for (auto& e : edge[si])
		{
			cnt[getDegree(e)]++;
			cnt[getDegree(e, 1)]--;
		}
		cnt[getDegree(si)]++;
	}

	if (res == 0)
		return -1;

	return res;
}

int k = 0;
int CountCritical()
{
	k++;
	for (int i = 0; i < n; i++)
	{
		order[i] = 0;
		par[i] = -1;
	}

	t = 0;

	vector<int> cx;
	for (int i = 0; i < n && cx.size() < 2; i++)
	{
		if (order[i] != 0)
			continue;

		int now = solve(i);

		if (now == 0)
			continue;

		if (now == -1)
			return 0;

		cx.push_back(now);
	}

	if (cx.size() >= 2)
		return 0;

	if (cx.empty())
		return n;

	return cx[0];
}

• Subtask 1: 20 / 20
• Subtask 2: 17 / 17
• Subtask 3: 0 / 18
• Subtask 4: 0 / 14
• Subtask 5: 0 / 31