Submission #948150

#	Time	Username	Problem	Language	Result	Execution time	Memory
948150		GrindMachine	Potemkin cycle (CEOI15_indcyc)	C++17	100 / 100	139 ms	91284 KiB

indcyc

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "yes" << endl
#define no cout << "no" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

read the solution a long time ago, remember some ideas from there

*/

const int MOD = 1e9 + 7;
const int N = 1e3 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
const int M = 2e5 + 5;

vector<pii> adj1[N];
vector<int> adj2[M];
bool edge[N][N];
vector<bool> vis(M), recstack(M);
vector<ll> curr_path;
vector<ll> cyc;

void dfs1(ll u){
    vis[u] = 1;
    recstack[u] = 1;
    curr_path.pb(u);

    trav(v,adj2[u]){
        if(vis[v]){
            if(recstack[v]){
                cyc = curr_path;
                reverse(all(cyc));
                while(cyc.back() != v){
                    cyc.pop_back();
                }
                reverse(all(cyc));
            }
        }
        else{
            dfs1(v);
        }
    }

    curr_path.pop_back();
    recstack[u] = 0;
}

void solve(int test_case)
{
    int n,m; cin >> n >> m;
    vector<pii> edges;
    rep1(u,n) edge[u][u] = 1;
    rep(i,m){
        int u,v; cin >> u >> v;
        adj1[u].pb({v,i*2}), adj1[v].pb({u,i*2+1});
        edges.pb({u,v}), edges.pb({v,u});
        edge[u][v] = edge[v][u] = 1;
    }

    rep1(u,n){
        for(auto [v,id1] : adj1[u]){
            for(auto [w,id2] : adj1[u]){
                if(edge[v][w]) conts;
                adj2[id1^1].pb(id2);
            }
        }
    }

    rep(i,2*m){
        if(vis[i]) conts;
        dfs1(i);
        if(!cyc.empty()) break;
    }

    if(cyc.empty()){
        no;
        return;
    }

    vector<ll> ans;
    ans.pb(edges[cyc[0]].ff);
    ans.pb(edges[cyc[0]].ss);
    rep1(i,sz(cyc)-2){
        assert(ans.back() == edges[cyc[i]].ff);
        ans.pb(edges[cyc[i]].ss);
    }

    trav(x,ans) cout << x << " ";
    cout << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10