Submission #1032026

#	Time	Username	Problem	Language	Result	Execution time	Memory
1032026		GrindMachine	Overtaking (IOI23_overtaking)	C++17	100 / 100	1456 ms	109644 KiB

overtaking

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

already knew some key ideas

*/

const int MOD = 1e9 + 7;
const int N = 1e3 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "overtaking.h"

vector<ll> T;
vector<int> W,S;
ll X,m;
vector<pll> segs;
vector<ll> fv;

void init(int L, int n, std::vector<long long> T_, std::vector<int> W_, int X_, int M, std::vector<int> S_)
{
    T = T_, W = W_, S = S_;
    X = X_, m = M;

    ll e[n][m], t[n][m];
    rep(i,n){
        e[i][0] = t[i][0] = T[i];
    }

    rep1(j,m-1){
        rep(i,n){
            e[i][j] = t[i][j-1]+(ll)W[i]*(S[j]-S[j-1]);
            t[i][j] = e[i][j];
        }

        vector<pll> ord;
        rep(i,n){
            ord.pb({t[i][j-1],i});
        }

        sort(all(ord));
        ll ptr = 0, mx = 0;

        rep(i,n){
            while(ptr < n and ord[ptr].ff < ord[i].ff){
                amax(mx,e[ord[ptr].ss][j]);
                ptr++;
            }

            amax(t[ord[i].ss][j],mx);
        }
    }

    vector<ll> b;
    rep(i,n){
        rep(j,m){
            b.pb(t[i][j]-X*S[j]);
        }
    }

    sort(all(b));
    b.resize(unique(all(b))-b.begin());

    segs.pb({-inf2,b[0]-1});
    rep(i,sz(b)){
        segs.pb({b[i],b[i]});
        if(i+1 < sz(b)){
            segs.pb({b[i]+1,b[i+1]-1});
        }
    }
    segs.pb({b.back()+1,inf2});
    ll siz = sz(segs);

    ll f[n][m];
    rep(i,n) f[i][m-1] = t[i][m-1];

    fv = vector<ll>(siz,-1);

    rev(j,m-2,0){
        vector<pll> ord;
        rep(i,n){
            ord.pb({t[i][j+1],i});
        }

        sort(all(ord));
        for(auto [tv,i] : ord){
            ll l = t[i][j]-X*S[j];
            ll r = t[i][j+1]-X*S[j+1];
            l = lower_bound(all(segs),pll{l,l})-segs.begin();
            r = lower_bound(all(segs),pll{r,r})-segs.begin();
            l++, r--;
            for(int ind = l; ind <= r; ++ind){
                fv[ind] = f[i][j+1];
            }
        }

        rep(i,n){
            ll val = t[i][j]-X*S[j];
            ll ind = lower_bound(all(segs),pll{val,val})-segs.begin();
            f[i][j] = max(t[i][j]+X*(S[m-1]-S[j]),fv[ind]);
        }
    }
}

long long arrival_time(long long t)
{
    ll ind = lower_bound(all(segs),pll{t+1,-1})-segs.begin()-1;
    ll ans = max(t+X*S[m-1],fv[ind]);
    return ans;
}

• Subtask 0: 0 / 0
• Subtask 1: 9 / 9
• Subtask 2: 10 / 10
• Subtask 3: 20 / 20
• Subtask 4: 26 / 26
• Subtask 5: 35 / 35