제출 #618170

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
618170		rrrr10000	Aliens (IOI16_aliens)	C++14	100 / 100	241 ms	7780 KiB

aliens

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
typedef tuple<ll,ll,ll> PP;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<bool> vb;
#define rep(i,n) for(ll i=0;i<(ll)(n);i++)
#define REP(i,k,n) for(ll i=(ll)(k);i<(ll)(n);i++)
#define all(a) a.begin(),a.end()
#define lb(v,k) (lower_bound(all(v),k)-v.begin())
#define fi first
#define se second
#define pb emplace_back
template<class T> void out(T a){cout<<a<<endl;}
template<class T> void outv(T v){rep(i,v.size()){if(i)cout<<' ';cout<<v[i];}cout<<endl;}
template<class T> bool chmin(T&a,T b){if(a>b){a=b;return true;}return false;}
template<class T> bool chmax(T&a,T b){if(a<b){a=b;return true;}return false;}
const ll inf=1001001001001001001;
struct CHT{
    deque<PP> deq;
    void insert(ll id,ll a,ll b){
        ll tmp;
        while(deq.size()>1){
            long double a1,b1,a2,b2;tie(a1,b1,tmp)=deq[deq.size()-2];tie(a2,b2,tmp)=deq.back();
            long double x=(b1-b)/(a-a1);
            if(a2*x+b2<a*x+b)break;
            deq.pop_back();
        }
        deq.pb(a,b,id);
    }
    P sol(ll x){
        while(deq.size()>1){
            if(get<0>(deq[0])*x+get<1>(deq[0])<get<0>(deq[1])*x+get<1>(deq[1]))break;
            deq.pop_front();
        }
        return P(get<0>(deq[0])*x+get<1>(deq[0]),get<2>(deq[0]));
    }
};
long long take_photos(int n, int m, int K, std::vector<int> r, std::vector<int> c) {
    vp srt(n);
    rep(i,n)srt[i]=P(max(r[i],c[i]),min(r[i],c[i]));
    sort(all(srt));
    vp v;v.pb(-1,-1);
    for(auto x:srt){
        while(v.size()&&v.back().fi>=x.se)v.pop_back();
        if(v.size()==0||v.back().se<x.fi)v.pb(x.se,x.fi);
    }
    n=v.size();chmin(K,n-1);
    auto sol=[&](ll lambda){
        vp dp(n);
        CHT cht;
        dp[0]=P(0,0);
        rep(i,n){
            if(i){
                auto res=cht.sol(v[i].se);
                dp[i].fi=res.fi+v[i].se*v[i].se+lambda;
                dp[i].se=dp[res.se].se+1;
            }
            if(i<n-1)cht.insert(i,2*(-v[i+1].fi+1),dp[i].fi+(-v[i+1].fi+1)*(-v[i+1].fi+1)-max(0ll,v[i].se-v[i+1].fi+1)*max(0ll,v[i].se-v[i+1].fi+1));
        }
        return dp[n-1];
    };
    if(K==n-1)return sol(0).fi;
    ll ok=(ll)(m)*(ll)(m)+5,ng=-1;
    while(ok-ng>1){
        ll md=(ok+ng)/2;
        auto res=sol(md);
        if(res.se<=K)ok=md;
        else ng=md;
    }
    P r1=sol(ok),r2=sol(ng);
    r1.fi-=r1.se*ok;r2.fi-=r2.se*ng;
    if(r1.se+r1.fi-(ng*(r2.se-r1.se)+r2.fi)>=K)return r1.fi-ok*(K-r1.se);
    return r2.fi+ng*(r2.se-K);
}

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