Submission #1184230

#	Time	Username	Problem	Language	Result	Execution time	Memory
1184230		asli_bg	Palembang Bridges (APIO15_bridge)	C++20	22 / 100	103 ms	14664 KiB

bridge

#include<bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;

#define int long long

typedef pair<int,int> pii;
typedef vector<pii> vii;
typedef vector<int> vi;
typedef vector<bool> vb;

typedef tree<pii,null_type,less<pii>,rb_tree_tag,
tree_order_statistics_node_update> oset;

#define FOR(i,a) for(int i=0;i<(a);i++)
#define FORE(i,a,b) for(int i=(a);i<(b);i++)

#define all(x) x.begin(),x.end()
#define fi first
#define se second
#define pb push_back
#define sp <<" "<<

#define cont(x) for(auto el:x) cout<<el<<' ';cout<<endl;
#define contp(x) for(auto el:x) cout<<el.fi<<'-'<<el.se<<' ';cout<<endl;

#define DEBUG(x) cout<<#x sp x<<endl;
#define carp(x,y) ((x%MOD)*(y%MOD))%MOD
#define topla(x,y) ((x%MOD)+(y%MOD))%MOD
#define mid (l+r)/2

const int MAXN=3e4+5;
const int MOD=1e9+7;
const int INF=1e18;

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int n,k;
    cin>>k>>n;
    vector<char> p(n+1), q(n+1);
    vi s(n+1), t(n+1);

    oset bir;

    int top,tane,ekle;
    top=tane=ekle=0;
    
    FORE(i,1,n+1){
        cin>>p[i]>>s[i]>>q[i]>>t[i];

        if(p[i]!=q[i]){
            bir.insert({t[i],n+i});
            bir.insert({s[i],i});

            top+=s[i]+t[i];
            tane+=2;
        }
        else ekle+=abs(t[i]-s[i]);
    }

    ekle+=tane/2;

    int res=INF;
    if(k<=2){
        int sz=bir.size();
        auto med=bir.find_by_order(sz/2);

        int deg=0;
        for(auto el:bir){
            if(el.fi>med->fi) deg+=el.fi-med->fi;
            else deg+=med->fi-el.fi; 
        }

        res=min(res,deg);
    }

    if(k==2){
        vii pref(2*n+1,{0,0});
        int once=-1;
        FOR(i,bir.size()){
            auto deg=*bir.find_by_order(i);
            pref[deg.se].fi=(once>=0?pref[once].fi:0)+deg.fi;
            pref[deg.se].se=(once>=0?pref[once].se:0)+1;

            once=deg.se;
        }

        vi tut(2*n+1);

        int ind=(int) bir.size()-1;

        int ekle2=0;
        int ekle3=0;

        oset iki;
        map<int,int> say;
        for(auto el:bir){
            say[(el.se<=n?el.se:el.se-n)]++;
            iki.insert(el);
        }

        while(!bir.empty()){
            int sz=bir.size();
            auto med=bir.find_by_order(sz/2);

            int tane1=pref[med->se].se;
            int tane2=tane-tane1;
            int top1=pref[med->se].fi;
            int top2=top-top1;

            tut[ind]+=(med->fi*tane1-top1)+(top2+ekle2-med->fi*(tane2+ekle3));

            auto deg=*bir.rbegin();
            top-=deg.fi;
            tane--;

            int ind2=deg.se;
            if(ind2>n) ind2-=n;

            if(say[ind2]==2){ //baş içerde son dışarda
                ekle2+=max(t[ind2],s[ind2]);
                ekle3++;
            }
            else if(say[ind2]==1){
                ekle2-=max(t[ind2],s[ind2]);
                ekle3--;
            }

            say[ind2]--;
            bir.erase(deg);
            ind--;
        }

        top=tane=0;

        ind=iki.size()-1;

        ekle2=0;
        ekle3=0;

        int dur=iki.size();
        
        while(dur!=bir.size()+1){
            auto deg=*iki.rbegin();

            int ind2=deg.se;
            if(ind2>n) ind2-=n;

            bir.insert(deg);

            int sz=bir.size();
            auto med=bir.find_by_order((sz-1)/2);

            if(say[ind2]==0){ //baş içerde son dışarda
                ekle2-=max(t[ind2],s[ind2]);
                ekle3--;
            }
            else if(say[ind2]==1){
                ekle2+=max(t[ind2],s[ind2]);
                ekle3++;
            }

            say[ind2]++;

            top+=deg.fi;
            tane++;

            int tane2=pref.back().se-pref[med->se].se;
            int tane1=tane-tane2;
            int top2=pref.back().fi-pref[med->se].fi;
            int top1=top-top2;

            tut[ind-1]+=(med->fi*tane1-top1)+(top2+ekle2-med->fi*(tane2+ekle3));

            iki.erase(deg);
            ind--;
        }

        FOR(i,n) res=min(res,tut[i]);
    }

    cout<<ekle+res<<endl;
} 

• Subtask 1: 8 / 8
• Subtask 2: 14 / 14
• Subtask 3: 0 / 9
• Subtask 4: 0 / 32
• Subtask 5: 0 / 37