Submission #477004

#	Time	Username	Problem	Language	Result	Execution time	Memory
477004		JvThunder	Keys (IOI21_keys)	C++17	100 / 100	1000 ms	51676 KiB

keys

#include <bits/stdc++.h>
#include "keys.h"
using namespace std;
 
const int maxn = 3e5 + 5;
int n,m,p[maxn],r[maxn];

int par[maxn];
int sz[maxn]; 
int find(int x) 
{ 
	if (par[x]==x) return x; 
	return par[x]=find(par[x]); 
}
 
bool have[maxn],vis[maxn],mk[maxn];
int hd,tl,q[maxn];
vector<pair<int,int>> g[maxn];
vector<int> vec[maxn];
vector<int> ans;
 
int bfs(int s,bool flag=0) 
{
	hd=1,tl=0; int res=0;
	q[++tl]=s,vis[s]=1;
	while (hd<=tl) 
	{
		int x=q[hd]; hd++;
		if (!have[r[x]]) 
		{
			while (!vec[r[x]].empty()) 
			{
				int y=vec[r[x]].back();
				if (find(y)!=s) res = find(y);
				if (!vis[y]) q[++tl] = y, vis[y] = 1;
				vec[r[x]].pop_back();
			}
			have[r[x]]=1;
		}
		for (pair<int,int> A : g[x]) 
		{
			int y=A.first;
			if (have[A.second]) 
			{
				if (find(y)!=s) res = find(y);
				if (!vis[y]) q[++tl] = y , vis[y] = 1;
			} 
			else vec[A.second].push_back(y);
		}
		if (res) break;
	}
	for (int i=1;i<=tl;i++) vis[q[i]] = 0;
	for (int i=1;i<hd;i++) 
	{
		int x = q[i];
		have[r[x]] = 0;
		for (pair<int,int> A : g[x]) 
			vec[A.second].clear();
	}
	if (flag)
	{
		for (int i=1;i<=tl;i++) 
			ans[q[i]-1] = tl;
	}
	return res;
}
 
vector<int> find_reachable(vector<int> R,vector<int> U,vector<int> V, vector<int> C) 
{
	n = R.size(); m = U.size();
	for (int i=0;i<n;i++) r[i+1] = R[i] + 1;
	int x,y,z; 
	ans = vector<int> (n,n+1);
	for (int i=0;i<m;i++) 
	{
		x=U[i]+1,y=V[i]+1,z=C[i]+1;
		g[x].push_back({y,z}),g[y].push_back({x,z});
	}
	for (int i=1;i<=n;i++) par[i] = i, sz[i] = 1;
	
	while (true) 
	{
		bool flag = false;
		for (int i=1;i<=n;i++) mk[i] = 0;
		for (int i=1;i<=n;i++) 
		{
			if (find(i)==i && !mk[i]) 
			{
				int t=bfs(i);
				if (t) par[i]=t,mk[t]=1,flag=1;
				else mk[i]=1;
			}
		}
		if (!flag) break;
	}
	int mn = n+1;
	for (int i=1;i<=n;i++) 
		if (find(i)==i) bfs(i,1);
	for (int i=1;i<=n;i++) mn = min(mn,ans[i-1]);
	for (int i=1;i<=n;i++) 
	{
		if (mn==ans[i-1]) ans[i-1] = 1; 
		else ans[i-1] = 0;
	}
	return ans;
}

• Subtask 1: 9 / 9
• Subtask 2: 11 / 11
• Subtask 3: 17 / 17
• Subtask 4: 30 / 30
• Subtask 5: 33 / 33