제출 #1078240

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1078240		c2zi6	추월 (IOI23_overtaking)	C++17	65 / 100	3557 ms	34964 KiB

overtaking

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "overtaking.h"

struct SEGTREE {
    int n;
    VL tree;
    SEGTREE(){}
    SEGTREE(int sz) {
        n = 1;
        while (n < sz) n *= 2;
        tree = VL(2*n);
    }
    void upd(int N, int L, int R, int i, ll s) {
        if (i < L || i > R) return;
        if (L == R) {
            tree[N] = s;
            return;
        }
        int M = (L + R) / 2;
        upd(2*N+1, L, M, i, s);
        upd(2*N+2, M+1, R, i, s);
        tree[N] = max(tree[2*N+1], tree[2*N+2]);
    }
    ll get(int N, int L, int R, int l, int r) {
        if (R < l || L > r) return 0;
        if (l <= L && R <= r) return tree[N];
        int M = (L + R) / 2;
        return max(get(2*N+1, L, M, l, r), get(2*N+2, M+1, R, l, r));
    }
    void upd(int i, ll s) {
        upd(0, 0, n-1, i, s);
    }
    ll get(int l, int r) {
        return get(0, 0, n-1, l, r);
    }
};

int n;
VI W, S;
VL T;
int m;

ll expected(int i, int j) {
    /* 
     * i-rd avtobusin inchqan jamanak e petq
     * (j-1)-rd stanciaic j-rd stancia hasnelu hamar
     */
    return 1ll * W[i] * (S[j] - S[j-1]);
}

typedef vector<tuple<ll, ll, int>> VLLI;

VLLI a;

VL step(VL t, int j) {
    int n = t.size();
    /*
     * t[i]-n i-rd avtobusi (j-1)-rd stancia
     * hasnelu jamanakn e, petq everadardznel ret,
     * urdex ret[i]-n i-rd avtobusi j-rd stancia
     * hasnelu jamanakn e
     */
    VL e(n);
    rep(i, n) e[i] = t[i] + expected(i, j);
    a = VLLI(n);
    rep(i, n) a[i] = {t[i], e[i], i};
    sort(all(a));
    ll maxexp = 0;
    VL ret(n);
    for (auto[t, e, i] : a) {
        setmax(maxexp, e);
        ret[i] = maxexp;
    }
    return ret;
}

struct BESTCHOICE {
    int n;
    VL times, values;
    SEGTREE seg;
    BESTCHOICE(){}
    BESTCHOICE(VLLI a) {
        n = a.size();
        times = values = VL();
        for (auto[t, e, i] : a) {
            times.pb(t);
            values.pb(e);
        }
        seg = SEGTREE(n);
        rep(i, n) seg.upd(i, values[i]);
    }
    ll operator()(ll x) {
        int ind = lower_bound(all(times), x) - times.begin() - 1;
        return seg.get(0, ind);
    }
    void print() {
        for (ll x : times) cout << x << " ";
        cout << endl;
        for (ll x : values) cout << x << " ";
        cout << endl;
    }
};

vector<BESTCHOICE> best;

void init(int L, int N_ARG, VL T_ARG, VI W_ARG, int X, int M, VI S_ARG) {
    n = N_ARG;
    W = W_ARG;
    W.pb(X);
    m = M;
    S = S_ARG;
    T = T_ARG;

    VL cur = T;
    replr(i, 1, m-1) {
        cur = step(cur, i);
        best.pb(BESTCHOICE(a));
    }
}

ll arrival_time(ll Y) {
    replr(i, 1, m-1) {
        auto& p = best[i-1];

        /*cout << "Kancinq hetevyal bani mijov" << endl;*/
        /*p.print();*/
        /*cout << "Expected@ " << Y + expected(n, i) << endl;*/

        Y = max(Y + expected(n, i), p(Y));

        /*cout << "Y dardzav " << Y << endl;*/
        /*cout << endl;*/
    }
    return Y;
}

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