제출 #653175

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
653175		perchuts	Two Antennas (JOI19_antennas)	C++17	100 / 100	995 ms	42412 KiB

antennas

//https://codeforces.com/blog/entry/66022?#comment-500622 

//line sweep com 4 tipos de eventos: 
//1-query 
//2-inserir antena 
//3-remover antena 
//4-atualizar a resposta 

//pra processar eventos, usar seg + lazy

//o porquê de precisar processar os eventos nas 2 ordens: note que quermos selecionar dois indices i<j pra cada query, e na seg,
//ou h[i]>h[j] ou h[j]>h[i]. Daí, queremos maximizar ou h[i] ou h[j]. Com os eventos crescendo no tempo maximizamos h[i] e maximizamos
//h[j] diminuindo no tempo. Dá pra fazer tudo de uma vez, mas o código ficaria mais confuso.
#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define sz(x) (int) x.size()
#define endl '\n'
#define pb push_back
#define _ ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

using namespace std;

using ll = long long;
using ull = unsigned long long;
using ii = pair<int,int>;
using iii = tuple<int,int,int>;

const int inf = 1e9+1;
const int mod = 1e9+7;
const int maxn = 2e5+100;

template<typename X, typename Y> bool ckmin(X& x, const Y& y) { return (y < x) ? (x=y,1):0; }
template<typename X, typename Y> bool ckmax(X& x, const Y& y) { return (x < y) ? (x=y,1):0; }

pair<int, ii> ant[maxn];

int n, ans[maxn], lazy[4*maxn], c[4*maxn], d[4*maxn];

ii qu[maxn];

void push(int i,int l,int r){
    if(lazy[i]==inf)return;

    ckmax(d[i], c[i] - lazy[i]);

    if(l!=r)ckmin(lazy[2*i], lazy[i]), ckmin(lazy[2*i+1], lazy[i]);

    lazy[i] = inf;
}

void update(int i,int l,int r,int x,int y,int k){
    if(x>y)return;
    
    push(i,l,r);

    if(l>y||r<x)return;
    
    if(x<=l&&r<=y){
        ckmin(lazy[i], k);
        push(i,l,r); 
        return;
    }

    int md = (l+r)/2;

    update(2*i, l, md, x, y, k), update(2*i+1, md+1, r, x, y, k);

    d[i] = max(d[2*i], d[2*i+1]), c[i] = max(c[2*i], c[2*i+1]);
}

int query(int i,int l,int r,int x,int y){
    push(i,l,r);

    if(l>y||r<x)return -1;

    if(x<=l&&r<=y)return d[i];

    int md = (l+r)/2;

    return max(query(2*i, l, md, x, y), query(2*i+1, md+1, r, x, y));
}

void upd(int i,int l,int r,int x,int k){
    push(i,l,r);

    if(l>x||r<x)return;

    if(l==r){
        c[i] = k;
        return;
    }   

    int md = (l+r)/2;

    upd(2*i, l, md, x, k), upd(2*i+1, md+1, r, x, k);

    d[i] = max(d[2*i], d[2*i+1]), c[i] = max(c[2*i], c[2*i+1]);
}

void init(){
    for(int i=0;i<=4*n;++i)d[i] = c[i] = -1, lazy[i] = inf;
}

int main(){_
    cin>>n;

    vector<pair<int,ii>>eventos1, eventos2;

    for(int i=0;i<n;++i){
        cin>>ant[i].first>>ant[i].second.first>>ant[i].second.second;
        
        eventos1.pb({i+ant[i].second.first, {-2, i}});
        eventos1.pb({i, {-1, i}});
        eventos1.pb({i+ant[i].second.second, {maxn, i}});

        eventos2.pb({i-ant[i].second.first, {maxn+1, i}});
        eventos2.pb({i, {maxn, i}});
        eventos2.pb({i-ant[i].second.second, {-1, i}});
    }

    int q;cin>>q;

    fill(ans,ans+q,-1);

    for(int i=0;i<q;++i){
        cin>>qu[i].first>>qu[i].second;
        --qu[i].first, --qu[i].second;
        eventos1.pb({qu[i].second,{i, qu[i].first}});
        eventos2.pb({qu[i].first,{i, qu[i].second}});
    }

    sort(all(eventos1)), sort(rbegin(eventos2), rend(eventos2));

    init();

    for(auto [x,p]:eventos1){
        auto [t, y] = p;
        if(t==-1){//atualizar o d (resposta)
            update(1,0,n-1, max(0, y-ant[y].second.second), y-ant[y].second.first, ant[y].first);
        }
        else if(t==-2){//inserir o y na seg
            upd(1,0,n-1, y, ant[y].first);                                          
        }
        else if(t==maxn){//remover o y da seg
            upd(1,0,n-1, y, -inf);
        }
        else{//responder a query de indice t
            ckmax(ans[t], query(1,0,n-1, y, x));
        }
    }

    init();

    for(auto [x,p]:eventos2){
        auto [t, y] = p;
        if(t==maxn){//atualizar o d (resposta)
            update(1,0,n-1, y+ant[y].second.first, min(n-1,y+ant[y].second.second), ant[y].first);
        }
        else if(t==maxn+1){//inserir o y na seg
            upd(1,0,n-1, y, ant[y].first);
        }
        else if(t==-1){//remover o y da seg
            upd(1,0,n-1, y, -inf);
        }
        else{//responder a query de indice t
            ckmax(ans[t], query(1,0,n-1, x, y));
        }
    }

    for(int i=0;i<q;++i)cout<<ans[i]<<endl;
}

• Subtask 1: 2 / 2
• Subtask 2: 11 / 11
• Subtask 3: 22 / 22
• Subtask 4: 65 / 65