Submission #412507

#	Time	Username	Problem	Language	Result	Execution time	Memory
412507		vlamarca	Aliens (IOI16_aliens)	C++17	100 / 100	221 ms	7852 KiB

aliens

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

#define fr(i,n) for(int i = 0; i<n; i++)
#define sz(v) (int)(v.size())
#define prin(a) cout << #a << " = " << a << endl
#define prinv(v) cout << #v << " = "; for(auto it : v) cout << it << ", "; cout << endl
#define all(v) (v).begin(),(v).end()

typedef long long ll;

#define rmin(a,b) a = min<ll>(a,b)
#define rmax(a,b) a = max<ll>(a,b)

#define fi first
#define se second

struct Line{
	// y = m*x + k
	ll m,k,qnt;
	ll eval(ll x){
		return m*x+k;
	}
	Line(){}
	Line(ll _m, ll _k, ll _qnt){
		m = _m, k = _k, qnt = _qnt;
	}
};

//querys em ordem de x decrescente
//insercoes de linhas com m (slopes) em ordem decrescente
struct ChtDeque{
	vector<Line> dq;
	int l, r; // [l,r)
	ChtDeque(){}
	ChtDeque(int n){ //n eh o maximo de linhas a serem inseridas
		n+=5;
		dq.resize(n+5);
		l = r = 3;
	}
	ll div(ll a, ll b){
		return a/b - ( (a^b)<0 and a%b);
	}
	bool tirar(Line &l1, Line &l2, Line &l3){
		if(l1.m==l2.m){
			return l1.k>=l2.k;
		}
		return div(l2.k-l3.k,l3.m-l2.m)<=div(l1.k-l2.k,l2.m-l1.m);
	}
	void add(ll m, ll k, ll qnt){
		m*=-1,k*=-1;
		Line line(m,k,qnt);
		while(r-l>=2){
			if(tirar(dq[r-2],dq[r-1],line)) r--;
			else break;
		}			
		if(l<r and dq[l].m==m){
			if(k>dq[l].k or (k==dq[l].k and qnt<dq[l].qnt)) dq[l] = line;
			return;
		}
		dq[r++] = line;
	}
	pair<ll,ll> query(ll x){
		assert(r>l);
		/*
		ll maxv = -LLONG_MAX, minq = 0;
		for(int i = l; i<r; i++){
			ll cur = dq[i].eval(x);
			ll curq = dq[i].qnt;
			if(cur>maxv or (cur==maxv and curq<minq)){
				maxv = cur;
				minq = curq;
			}
		}
		return {-maxv,minq};
		*/
		while(r-l>=2){
			ll val_l = dq[l].eval(x);
			ll val_l1 = dq[l+1].eval(x);
			if(val_l1>val_l) l++;
			else if(val_l1==val_l){
				if(dq[l].qnt<dq[l+1].qnt) break;
				else l++;
			}
			else break;			
		}
		return {-dq[l].eval(x),dq[l].qnt};
	}
};

vector<pair<ll,ll>> vp;
int debug = 0;
//ans min, k used
pair<ll,ll> f(ll c){
	ChtDeque cht(sz(vp)+1);
	ll udp = 0, uqnt = 0;
	fr(i,sz(vp)){
		ll ri,ci; tie(ri,ci) = vp[i];
		ci++;
		ll sub = 0;
		if(i){
			ll rj,cj; tie(rj,cj) = vp[i-1];
			cj++;
			if(cj>ri) sub = (cj-ri)*(cj-ri); 
		}
		cht.add(-2*ri,ri*ri+udp+c-sub,uqnt);
		
		ll dp, qnt; tie(dp,qnt) = cht.query(ci);
		dp+=ci*ci;
		qnt++;
		udp = dp, uqnt = qnt;
	}
	return {udp,uqnt};
}

ll take_photos(int n, int m, int k, vector<int> vr, vector<int> vc){
	vp.clear();
	fr(i,n){
		if(vc[i]<vr[i]) swap(vc[i],vr[i]);
		vp.emplace_back(vr[i],vc[i]);
	}
	{
		sort(all(vp));
		reverse(all(vp));
		vector<pair<ll,ll>> aux;
		for(auto &[i,j] : vp){
			if(!aux.empty() and aux.back().fi==i) continue;
			while(!aux.empty() and j>=aux.back().se) aux.pop_back();
			aux.emplace_back(i,j);
		}
		reverse(all(aux));
		vp = aux;
	}
	ll lo = 0, hi = (ll)m*m;
	while(lo<hi){
		ll mid = (lo+hi)/2;
		if(f(mid).se<=k) hi = mid;
		else lo = mid+1;
	};
	auto par = f(lo);
	/*
	prin(par.fi);
	prin(k);
	prin(par.se);
	prin(lo);*/
	return par.fi-(ll)k*lo;
}

• Subtask 1: 4 / 4
• Subtask 2: 12 / 12
• Subtask 3: 9 / 9
• Subtask 4: 16 / 16
• Subtask 5: 19 / 19
• Subtask 6: 40 / 40