Submission #439985

#	Time	Username	Problem	Language	Result	Execution time	Memory
439985		haojiandan	Keys (IOI21_keys)	C++17	100 / 100	1222 ms	51568 KiB

keys

#include <bits/stdc++.h>
#include "keys.h"
using namespace std;
#define MP make_pair
const int maxn=(3e5)+10;
int n,m,p[maxn],r[maxn],fa[maxn];
int sz[maxn];
int find(int x) { if (fa[x]==x) return x; return fa[x]=find(fa[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=(int)R.size();
	for (int i=0;i<n;i++) r[i+1]=R[i]+1;
	m=(int)U.size(); 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(MP(y,z)),g[y].push_back(MP(x,z));
	}
	for (int i=1;i<=n;i++) fa[i]=i,sz[i]=1;
	
	while (1) {
		bool flag=0;
		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) fa[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=0;i<n;i++) printf("%d ",ans[i]); puts("");
	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