Submission #527507

#	Time	Username	Problem	Language	Result	Execution time	Memory
527507		zaneyu	Parkovi (COCI22_parkovi)	C++14	110 / 110	1287 ms	54152 KiB

Main

/*input
2 1 
1 2 3
*/ 
#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=1e6+5;
vector<pii> v[maxn];
ll k;
vector<int> anss;
pii dfs(int u,int p,ll pv){
    ll mx=0,mx2=-INF64;
    for(auto x:v[u]){
        if(x.f!=p){
            pii z=dfs(x.f,u,x.s);
            if(!z.f){
                MXTO(mx,z.s);
            }
            else{
                MXTO(mx2,z.s);
            }
        }
    }
    pii ans;
    if(mx2>=mx){
        ans.f=1;
        ans.s=(mx2-pv);
    }
    else{
        ans.f=0;
        ans.s=(mx+pv);
    }
    if(!ans.f and ans.s>k){
       	anss.pb(u);
        ans.f=1;
        ans.s=k-pv;
    }
    if(ans.f and ans.s<0) ans.f=0,ans.s=0;
    return ans;
}
bool in[maxn];
int main(){
    int n,z;
    cin>>n>>z;
    REP(i,n-1){
        int a,b,c;
        cin>>a>>b>>c;
        --a;
        --b;
        v[a].pb({b,c});
        v[b].pb({a,c});
    }
    ll l=0,r=INF64;
    while(l<r){
        ll mid=(l+r)/2;
        k=mid;
        dfs(0,-1,INF64);
        if(sz(anss)<=z) r=mid;
        else l=mid+1;
        anss.clear();
    }
    k=l;
    dfs(0,-1,INF64);
    cout<<l<<'\n';
    for(int x:anss) in[x]=1;
    int p=0;
    while(sz(anss)<z){
    	while(in[p]) ++p;
    	anss.pb(p);
    	++p;
    }
    cout<<anss;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 30 / 30
• Subtask 4: 60 / 60