답안 #990045

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
990045 2024-05-29T12:14:35 Z PedroBigMan Training (IOI07_training) C++17
0 / 100
4 ms 1884 KB
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("Ofast")
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <queue>
#include <deque>
#include <list>
#include <iomanip>
#include <stdlib.h>
#include <time.h>
#include <cstring>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
#define REP(i,a,b) for(ll i=(ll) a; i<(ll) b; i++)
#define pb push_back
#define mp make_pair
#define pl pair<ll,ll>
#define ff first
#define ss second
#define whole(x) x.begin(),x.end()
#define DEBUG(i) cout<<"Pedro "<<i<<endl
#define INF 1000000000000000000LL
#define EPS ((ld)0.00000000001)
#define pi ((ld)3.141592653589793)
#define VV(vvvv,NNNN,xxxx); REP(iiiii,0,NNNN) {vvvv.pb(xxxx);}
ll mod=1000000007;

template<class A=ll> 
void Out(vector<A> a) {REP(i,0,a.size()) {cout<<a[i]<<" ";} cout<<endl;}

template<class A=ll>
void In(vector<A> &a, ll N) {A cur; REP(i,0,N) {cin>>cur; a.pb(cur);}} 

class ST
{	
    public:
    ll N;
    
    class SV //seg value
    {
        public:
        ll a; 
        SV() {a=0LL;}
        SV(ll x) {a=x;}
        
        SV operator & (SV X) {SV ANS(a+X.a); return ANS;}
    };
      
    class LV //lazy value
    {
        public:
        ll a;
        LV() {a=0LL;}
        LV(ll x) {a=x;}
        
        LV operator & (LV X) {LV ANS(a+X.a); return ANS;}
    };
    
    SV upval(ll c) //how lazy values modify a seg value inside a node, c=current node
    {
        SV X(p[c].a+(range[c].ss-range[c].ff+1)*lazy[c].a);
        return X;
    }
    
    SV neuts; LV neutl;
    
    vector<SV> p;
    vector<LV> lazy;
    vector<pl> range;
    
    ST() {N=0LL;}
    ST(vector<ll> arr)
    {
        N = (ll) 1<<(ll) ceil(log2(arr.size()));
        REP(i,0,2*N) {range.pb(mp(0LL,0LL));}
        REP(i,0,N) {p.pb(neuts);}
        REP(i,0,arr.size()) {SV X(arr[i]); p.pb(X); range[i+N]=mp(i,i);}
        REP(i,arr.size(),N) {p.pb(neuts); range[i+N]=mp(i,i);}
        ll cur = N-1;
        while(cur>0)
        {
            p[cur]=p[2*cur]&p[2*cur+1];
            range[cur]=mp(range[2*cur].ff,range[2*cur+1].ss);
            cur--;
        }
        REP(i,0,2*N) {lazy.pb(neutl);}
    }
    
    void prop(ll c) //how lazy values propagate
    {
        p[c] = upval(c);
        lazy[2*c]=lazy[c]&lazy[2*c]; lazy[2*c+1]=lazy[c]&lazy[2*c+1];
        lazy[c]=neutl;
    }
    
    SV query(ll a,ll b, ll c=1LL) //range [a,b], current node. initially: query(a,b)
    {
		if(a>b) {return neuts;}
        ll x=range[c].ff; ll y=range[c].ss;
        if(y<a || x>b) {return neuts;}
        if(x>=a && y<=b) {return upval(c);}
        prop(c);
        SV ans = query(a,b,2*c)&query(a,b,2*c+1);
        return ans;
    }
    
    void update(LV s, ll a, ll b, ll c=1LL) //update LV, range [a,b], current node, current range. initially: update(s,a,b)
    {
		if(a>b) {return;}
        ll x=range[c].ff; ll y=range[c].ss;
        if(y<a || x>b) {return ;}
        if(x>=a && y<=b) 
        {
            lazy[c]=s&lazy[c]; 
            return;
        }
		prop(c);
        update(s,a,b,2*c); update(s,a,b,2*c+1);
        p[c]=upval(2*c)&upval(2*c+1);
    }
};

template<class T=ll>
class SparseTable //Range Minimum Queries
{
    public:
    ll N; 
    vector<T> a;
    vector<vector<T> > v;
    
    SparseTable() {N=0LL;}
    SparseTable(vector<T> b)
    {
        a=b; N=a.size();
        ll lo=(ll) floor((ld) log2(N)) +1LL;
        vector<T> xx; 
        REP(i,0,lo) {xx.pb(mp(INF,INF));} REP(i,0,N) {v.pb(xx);}
        REP(step,0LL,lo)
        {
            REP(i,0,N-(1LL<<step)+1LL)
            {
                if(step==0) {v[i][0]=a[i];}
                else {v[i][step]=min(v[i][step-1],v[i+(1LL<<(step-1))][step-1]);}
            }
        }
    }
    
    T query(ll l, ll r)
    {
        ll step=(ll) floor((ld) log2(r-l+1LL));
        return min(v[l][step],v[r-(1LL<<step)+1LL][step]);
    }
};

class Tree
{
    public:
    ll N; 
    vector<ll> p; 
    vector<vector<ll> > sons;
    vector<vector<ll> > adj;
    ll root;
    vector<ll> level; //starting in 0
    vector<ll> val; //node values
    vector<ll> DFSarr1; //DFS Array
    vector<ll> DFSarr2; //DFS Array for LCA with whole path
    vector<ll> pos; //inverted DFSArr, only for LCA
    vector<pl> levDFSarr; //array of levels on DFSarr, only needed for LCA
    vector<pl> range; vector<ll> pos1;
    SparseTable<pl> S; //for LCA
    vector<ll> dp0, dp1;
    ST F;
    vector<vector<pair<pl,ll> > > paths;

    Tree(vector<vector<ll> > ad, ll r=0LL)
    {
        N=ad.size(); root=r; adj=ad;
        vector<ll> xx; REP(i,0,N) {sons.pb(xx); p.pb(-1); level.pb(0); pos.pb(0LL); val.pb(0); dp0.pb(0); dp1.pb(0); range.pb({-1,-1}); pos1.pb(-1);}
        DFS_Build(r,r);
        REP(i,0,DFSarr2.size()) {pos[DFSarr2[i]]=i;}
        REP(i,0,DFSarr2.size()) {levDFSarr.pb(mp(level[DFSarr2[i]],DFSarr2[i]));}
        SparseTable<pl> X(levDFSarr); S=X;
        F = ST(val);
        paths = vector<vector<pair<pl,ll> > >(N,vector<pair<pl,ll> >());
    }
        
    void DFS_Build(ll s, ll par)
    { 
        DFSarr1.pb(s); pos1[s]=DFSarr1.size()-1; range[s].ff=pos1[s];
        DFSarr2.pb(s);
        if(s!=root) {level[s]=level[par]+1LL;}
        p[s]=par;
        REP(i,0,adj[s].size())
        {
            if(adj[s][i]==par) {continue;}
            sons[s].pb(adj[s][i]);
            DFS_Build(adj[s][i],s);
            DFSarr2.pb(s);
        }
        range[s].ss = DFSarr1.size()-1;
        return;
    }
           
    ll LCA(ll a, ll b)
    {
        a=pos[a]; b=pos[b]; 
        ll l=min(a,b); ll r=max(a,b);
        pl ans=S.query(l,r);
        return ans.ss;
    }

    ll query(ll a, ll b)
    {
        ll lca = LCA(a,b);
        a=pos1[a]; b=pos1[b]; lca=pos1[lca];
        return (F.query(a,a).a+F.query(b,b).a-F.query(lca,lca).a);
    }

    void update(ll s, ll x)
    {
        F.update(x,range[s].ff,range[s].ss);
    }

    void Init_paths(vector<pair<pl,ll> > p)
    {
        REP(i,0,p.size())
        {
            paths[LCA(p[i].ff.ff,p[i].ff.ss)].pb(p[i]);
        }
    }

    void DP(ll s)
    {
        REP(i,0,sons[s].size()) {DP(sons[s][i]);}
        ll x = 0LL; REP(i,0,sons[s].size()) {x+=dp1[sons[s][i]];}
        dp0[s]=x; update(s,x);
        dp1[s]=dp0[s];
        ll A, B, C;
        REP(i,0,paths[s].size())
        {
            A = paths[s][i].ff.ff; B = paths[s][i].ff.ss; C = paths[s][i].ss;
            dp1[s]=max(dp1[s],C+query(A,B));
        }
        update(s,-dp1[s]);
    }
};

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
	cout.precision(20);
	ll N,M; cin>>N>>M;
    vector<pair<pl,ll> > old_paths, paths; vector<vector<ll> > adj(N,vector<ll>());
    ll A,B,C; ll ans = 0LL;
    REP(i,0,M)
    {
        cin>>A>>B>>C; A--; B--; ans+=C;
        if(C==0) {adj[A].pb(B); adj[B].pb(A);}
        else {old_paths.pb({{A,B},C});}
    }  
    Tree T(adj);
    REP(i,0,old_paths.size())
    {
        A=old_paths[i].ff.ff; B=old_paths[i].ff.ss; C=old_paths[i].ss; 
        if((T.level[A]+T.level[B]+1LL)%2LL == 0) {continue;}
        paths.pb(old_paths[i]);
    }
    T.Init_paths(paths);
    T.DP(0);
    cout<<ans-T.dp1[0]<<endl;
    //Out(T.dp0); Out(T.dp1);
    return 0;
}

Compilation message

training.cpp:1: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    1 | #pragma GCC optimization ("O3")
      | 
training.cpp:2: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    2 | #pragma GCC optimization ("unroll-loops")
      |
# 결과 실행 시간 메모리 Grader output
1 Incorrect 0 ms 348 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 348 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 1884 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 348 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 0 ms 460 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 348 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 1 ms 860 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 1116 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 3 ms 1884 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 2 ms 1112 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 4 ms 1880 KB Output isn't correct
2 Halted 0 ms 0 KB -