Submission #1214805

#	Time	Username	Problem	Language	Result	Execution time	Memory
1214805		biank	Newspapers (CEOI21_newspapers)	C++20	100 / 100	36 ms	4676 KiB

newspapers

#include <bits/stdc++.h>

using namespace std;

#define forn(i,n) for(int i=0;i<int(n);i++)
#define forsn(i,s,n) for(int i=int(s);i<int(n);i++)
#define dforn(i,n) for(int i=int(n)-1;i>=0;i--)
#define dforsn(i,s,n) for(int i=int(n)-1;i>=int(s);i--)
#define fst first
#define snd second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()

typedef long long ll;
typedef vector<ll> vll;
typedef vector<int> vi;
typedef pair<int,int> ii;

const int INF=1e9;

const int MAX_N=1005;

int maxDepth[MAX_N];
vi adj[MAX_N];

int dfs0(int u, int p=-1){
    int maxi=0;
    for(int v:adj[u]) if(v!=p){
        maxi=max(maxi,dfs0(v,u)+1);
    }
    return maxDepth[u]=maxi;
}

int dist[MAX_N],par[MAX_N];

void dfs(int u){
    for(int v:adj[u]) if(v!=par[u]){
        par[v]=u;
        dist[v]=dist[u]+1;
        dfs(v);
    }
}

bool check(int n, int m){
    if(m!=n-1) return false;
    forn(u,n){
        int cnt=0;
        for(int v:adj[u]){
            if(dfs0(v,u)>=2) cnt++;
        }
        if(cnt>2) return false;
    }
    return true;
}

vi findDiameter(int n){
    par[0]=-1,dist[0]=0,dfs(0);
    int u=0;
    forsn(v,1,n) if(dist[v]>dist[u]) u=v;
    par[u]=-1,dist[u]=0,dfs(u);
    int u2=0;
    forsn(v,1,n) if(dist[v]>dist[u2]) u2=v;
    vi diameter{u2};
    while(diameter.back()!=u){
        diameter.pb(par[diameter.back()]);
    }
    return diameter;
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int n,m;
    cin>>n>>m;
    forn(i,m){
        int u,v;
        cin>>u>>v;
        --u,--v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    if(!check(n,m)){
        cout<<"NO\n";
        return 0;
    }
    cout<<"YES\n";
    if(n<=2){
        cout<<n<<'\n';
        forn(i,n) cout<<"1 ";
        cout<<'\n';
        return 0;
    }
    vi d=findDiameter(n);
    vi ret;
    forsn(i,1,sz(d)-1){
        int u=d[i];
        ret.pb(u);
        for(int v:adj[u]) if(sz(adj[v])!=1&&v!=d[i+1]&&v!=d[i-1]){
            ret.pb(v);
            ret.pb(u);
        }
    }
    cout<<2*sz(ret)<<'\n';
    forn(i,sz(ret)) cout<<ret[i]+1<<' ';
    dforn(i,sz(ret)) cout<<ret[i]+1<<' ';
    cout<<'\n';
    
    return 0;
}

• Subtask 1: 12 / 12
• Subtask 2: 8 / 8
• Subtask 3: 80 / 80