Submission #618978

#TimeUsernameProblemLanguageResultExecution timeMemory
618978CSQ31Team Contest (JOI22_team)C++17
0 / 100
122 ms30864 KiB
#include <bits/stdc++.h> using namespace std; #define pb push_back #define fi first #define se second #define sz(a) (int)(a.size()) #define all(a) a.begin(),a.end() #define owo ios_base::sync_with_stdio(0);cin.tie(0); typedef pair<int,int> pii; typedef long long int ll; const int MAXN = 5e5; array<ll,3>pts1[MAXN],pts2[MAXN]; vector<array<ll,3>>a; vector<int>crd[3]; //i have rectangle queries but i cant do it offline? //i can but its double log on 2s //ok try n^2 first int sz0,sz1,sz2; struct seg{ vector<int>t; int n; seg(){} seg(int _n){ n = _n; t.assign(4*n+1,-2e9); } void upd(int v,int pos,int l,int r,int val){ if(l==r){ t[v] = max(t[v],val); return; } int tm = (l+r)/2; if(pos<=tm)upd(2*v,pos,l,tm,val); else upd(2*v+1,pos,tm+1,r,val); t[v] = max(t[2*v],t[2*v+1]); } void upd2(int v,int pos,int l,int r,int val){ if(l==r){ t[v] = val; return; } int tm = (l+r)/2; if(pos<=tm)upd(2*v,pos,l,tm,val); else upd(2*v+1,pos,tm+1,r,val); t[v] = max(t[2*v],t[2*v+1]); } int query(int v,int l,int r,int tl,int tr){ if(l>r)return -2e9; if(l==tl && r==tr)return t[v]; int tm = (tl+tr)/2; if(r<=tm)return query(2*v,l,r,tl,tm); else if(l>tm)return query(2*v+1,l,r,tm+1,tr); else return max(query(2*v,l,tm,tl,tm),query(2*v+1,tm+1,r,tm+1,tr)); } }amx; ll ans = -1; vector<int>pts[MAXN]; void solve(int l,int r){ if(l==r){ return; } //all same a int mid = (l+r)/2; solve(l,mid); solve(mid+1,r); vector<pii>que; vector<array<ll,3>>add; for(int i=l;i<=mid;i++){ for(int x:pts[i]){ if(pts1[x][0] != -2e9){ add.pb({pts1[x][1],pts1[x][2],pts1[x][0]}); } if(pts2[x][0] != -2e9){ add.pb({pts2[x][1],pts2[x][2],pts2[x][0]}); } } } for(int i=mid+1;i<=r;i++){ for(int x:pts[i])que.push_back({a[x][1],x}); } sort(all(add),greater<array<ll,3>>()); sort(all(que),greater<pii>()); int ptr = -1; for(pii qq : que){ while(ptr+1 < sz(add) && add[ptr+1][0] > qq.fi){ //add still has bigger b ptr++; array<ll,3>v = add[ptr]; amx.upd(1,v[1],0,sz2-1,v[2]); } int id = qq.se; if(a[id][2]+1 < sz2){ ll res = amx.query(1,a[id][2]+1,sz2-1,0,sz2-1); //we need a bigger ccout<<"query"<<'\n'; ans = max(ans , crd[0][a[id][0]] + res); } } for(auto x:add)amx.upd(1,x[1],0,sz2-1,-2e9); } int main() { owo int n; cin>>n; a.resize(n); for(int i=0;i<n;i++){ for(int j=0;j<3;j++){ cin>>a[i][j]; crd[j].pb(a[i][j]); } } sort(all(a),[&](array<ll,3>i,array<ll,3>j){return j[0] > i[0];}); for(int j=0;j<3;j++){ sort(all(crd[j])); crd[j].resize(unique(all(crd[j])) - crd[j].begin()); } //cout<<"sorted"<<'\n'; //for(int i=0;i<n;i++)cout<<a[i][0]<<" "<<a[i][1]<<" "<<a[i][2]<<'\n'; sz0 = sz(crd[0]); sz1 = sz(crd[1]); sz2 = sz(crd[2]); seg bmx(sz1),cmx(sz2); //cout<<"points"<<'\n'; for(int i=0;i<n;i++){ pts1[i][0] = pts2[i][0] = -2e9; pts1[i][1] = a[i][1]; //i has bigger b pts2[i][2] = a[i][2]; //i has bigger c int s1 = lower_bound(all(crd[1]),a[i][1]) - crd[1].begin(); int s2 = lower_bound(all(crd[2]),a[i][2]) - crd[2].begin(); if(s1>0)pts1[i][2] = bmx.query(1,0,s1-1,0,sz1-1); //want the max of c among smaller bs if(s2>0)pts2[i][1] = cmx.query(1,0,s2-1,0,sz2-1); //want the max of b among smaller cs if(pts1[i][2] > a[i][2])pts1[i][0] = pts1[i][1] + pts1[i][2]; if(pts2[i][1] > a[i][1])pts2[i][0] = pts2[i][1] + pts2[i][2]; bmx.upd(1,s1,0,sz1-1,a[i][2]); cmx.upd(1,s2,0,sz2-1,a[i][1]); if(pts1[i][0] != -2e9){ pts1[i][1] = lower_bound(all(crd[1]),pts1[i][1]) - crd[1].begin(); pts1[i][2] = lower_bound(all(crd[2]),pts1[i][2]) - crd[2].begin(); } if(pts2[i][0] != -2e9){ pts2[i][1] = lower_bound(all(crd[1]),pts2[i][1]) - crd[1].begin(); pts2[i][2] = lower_bound(all(crd[2]),pts2[i][2]) - crd[2].begin(); } //this is correct } for(int i=0;i<n;i++){ a[i][0] = lower_bound(all(crd[0]),a[i][0]) - crd[0].begin(); a[i][1] = lower_bound(all(crd[1]),a[i][1]) - crd[1].begin(); a[i][2] = lower_bound(all(crd[2]),a[i][2]) - crd[2].begin(); pts[a[i][0]].pb(i); } /* for(int i=0;i<n;){ int k = i; while(k+1<n && a[k][0] == a[k+1][0])k++; for(int x=i;x<=k;x++){ for(int j=0;j<x;j++){ if(a[x][0] == a[j][0])break; if(pts1[j][1] > a[x][1] && pts1[j][2] >a[x][2])ans = max(ans,crd[0][a[x][0]] + pts1[j][0]); if(pts2[j][1] > a[x][1] && pts2[j][2] >a[x][2])ans = max(ans,crd[0][a[x][0]] + pts2[j][0]); } } i = k+1; }*/ amx = seg(sz2); solve(0,sz0-1); cout<<ans; }
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