Submission #1032036

#	Time	Username	Problem	Language	Result	Execution time	Memory
1032036		GrindMachine	Overtaking (IOI23_overtaking)	C++17	100 / 100	2037 ms	154836 KiB

overtaking

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")

#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"

template<typename T>
struct lazysegtree {
    /*=======================================================*/

    struct data {
        ll a;
    };

    struct lazy {
        ll a;
    };

    data d_neutral = {-1};
    lazy l_neutral = {-1};

    void merge(data &curr, data &left, data &right) {
        curr.a = max(left.a,right.a);
    }

    void create(int x, int lx, int rx, T v) {

    }

    void modify(int x, int lx, int rx, T v) {
        lz[x].a = v;
    }

    void propagate(int x, int lx, int rx) {
        ll v = lz[x].a;
        if(v == -1) return;

        tr[x].a = v;

        if(rx-lx > 1){
            lz[x<<1].a = v;
            lz[x<<1|1].a = v;
        }

        lz[x] = l_neutral;
    }

    /*=======================================================*/

    int siz = 1;
    vector<data> tr;
    vector<lazy> lz;

    lazysegtree() {

    }

    lazysegtree(int n) {
        while (siz < n) siz *= 2;
        tr.assign(2 * siz, d_neutral);
        lz.assign(2 * siz, l_neutral);
    }

    void build(vector<T> &a, int n, int x, int lx, int rx) {
        if (rx - lx == 1) {
            if (lx < n) {
                create(x, lx, rx, a[lx]);
            }

            return;
        }

        int mid = (lx + rx) >> 1;

        build(a, n, x<<1, lx, mid);
        build(a, n, x<<1|1, mid, rx);

        merge(tr[x], tr[x<<1], tr[x<<1|1]);
    }

    void build(vector<T> &a, int n) {
        build(a, n, 1, 0, siz);
    }

    void rupd(int l, int r, T v, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return;
        if (lx >= l and rx <= r) {
            modify(x, lx, rx, v);
            propagate(x, lx, rx);
            return;
        }

        int mid = (lx + rx) >> 1;

        rupd(l, r, v, x<<1, lx, mid);
        rupd(l, r, v, x<<1|1, mid, rx);

        merge(tr[x], tr[x<<1], tr[x<<1|1]);
    }

    void rupd(int l, int r, T v) {
        rupd(l, r + 1, v, 1, 0, siz);
    }

    data query(int l, int r, int x, int lx, int rx) {
        propagate(x, lx, rx);

        if (lx >= r or rx <= l) return d_neutral;
        if (lx >= l and rx <= r) return tr[x];

        int mid = (lx + rx) >> 1;

        data curr;
        data left = query(l, r, x<<1, lx, mid);
        data right = query(l, r, x<<1|1, mid, rx);

        merge(curr, left, right);
        return curr;
    }

    data query(int l, int r) {
        return query(l, r + 1, 1, 0, siz);
    }
};

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

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);
    st = lazysegtree<ll>(siz);

    ll f[n][m];
    rep(i,n) f[i][m-1] = t[i][m-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--;
            st.rupd(l,r,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]),st.query(ind,ind).a);
        }
    }
}

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],st.query(ind,ind).a);
    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