Submission #527081

#	Time	Username	Problem	Language	Result	Execution time	Memory
527081		zaneyu	Potemkin cycle (CEOI15_indcyc)	C++14	0 / 100	44 ms	1972 KiB

indcyc

/*input
5 6
1 2
1 3
2 3
4 3
5 2
4 5
*/ 
#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;

//#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
using ll = long long;
using ld = long double;
using pii = pair<ll,ll>;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define MP make_pair
const ll INF64=4e17;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]+1<<((i!=sz(V)-1)?" ":"");
    return out;
}
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=mult(res,a);
        a=mult(a,a);
        b>>=1;
    }
    return res;
}
const int maxn=1e3+5;
vector<int> v[maxn];
int vis[maxn];
void dfs(int u,vector<int> &vv){
    vv.pb(u);
    if(sz(vv)>=4){
        cout<<vv;
        exit(0);
    }
    vis[u]=1;
    for(int x:v[u]){
        if(!vis[x]){
            dfs(x,vv);
        }
    }
    vv.pop_back();
}
int main(){
    int n,m;
    cin>>n>>m;
    REP(i,m){
        int a,b;
        cin>>a>>b;
        --a,--b;
        v[a].pb(b),v[b].pb(a);
    }
    REP(i,n){
        REP(j,n) vis[j]=0;
        vector<int> vv;
        dfs(i,vv);
    }
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10