이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//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=1;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, {n, i}});
eventos2.pb({i-ant[i].second.first, {n+1, i}});
eventos2.pb({i, {n, 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==n){//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==n){//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==n+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;
}
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