Submission #1245992

#TimeUsernameProblemLanguageResultExecution timeMemory
1245992LittleOrangeNile (IOI24_nile)C++20
100 / 100
86 ms18988 KiB
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr ll big = 1e18;
struct dsu{
    ll n,ans;
    vector<ll> p,sm,cls;
    array<vector<ll>,2> mi;
    dsu(const vector<ll> &a):n(a.size()),ans(0),p(a.size(),-1),sm(a){
        cls.resize(n);
        iota(cls.begin(),cls.end(),0);
        mi[0].assign(n,big);
        mi[1].assign(n,big);
        for(ll i = 0;i<n;i++){
            mi[i&1][i] = a[i];
        }
    }
    ll g(ll i){
        return p[i]<0?i:p[i]=g(p[i]);
    }
    ll cal(ll i){
        return sm[i]-(p[i]&1)*mi[cls[i]&1][i];
    }
    void m(ll a, ll b){
        a = g(a),b = g(b);
        if(a==b)return;
        ans -= cal(a)+cal(b);
        if(p[a]>p[b]) swap(a,b);
        sm[a] += sm[b];
        mi[0][a] = min(mi[0][a],mi[0][b]);
        mi[1][a] = min(mi[1][a],mi[1][b]);
        cls[a] = min(cls[a],cls[b]);
        p[a] += p[b];
        p[b] = a;
        ans += cal(a);
    }
    void upd(ll i, ll v){
        ll typ = i&1;
        i = g(i);
        ans -= cal(i);
        mi[typ^1][i] = min(mi[typ^1][i],v);
        ans += cal(i);
    }
};
vector<long long> calculate_costs(vector<int> w, vector<int> a, vector<int> b, vector<int> e){
    ll n = w.size();
    ll q = e.size();
    ll tot = 0;
    for(ll i : a) tot += i;
    vector<pair<ll,ll>> v;
    for(ll i = 0;i<n;i++) v.push_back({w[i],a[i]-b[i]});
    sort(v.begin(),v.end());
    vector<ll> h(n);
    for(ll i = 0;i<n;i++) h[i] = v[i].second;
    dsu d(h);
    vector<pair<ll,ll>> u;
    for(ll i = 0;i<n-1;i++) u.push_back({v[i+1].first-v[i].first,i});
    for(ll i = 0;i<n-2;i++) u.push_back({v[i+2].first-v[i].first,-(i+1)});
    sort(u.begin(),u.end());
    vector<pair<ll,ll>> o;
    o.push_back({0,0});
    for(auto [dd,i] : u){
        if(i<0){
            d.upd(-i,h[-i]);
        }else{
            d.m(i,i+1);
        }
        o.push_back({dd,d.ans});
    }
    vector<ll> ret(q);
    for(ll i = 0;i<q;i++){
        ret[i] = tot-(*--lower_bound(o.begin(),o.end(),pair<ll,ll>((ll)e[i]+1,0ll))).second;
    }
    return ret;
}
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