Submission #1078075

#	Time	Username	Problem	Language	Result	Execution time	Memory
1078075		c2zi6	Overtaking (IOI23_overtaking)	C++17	39 / 100	3556 ms	600 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"

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]);
}

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);

    vector<tuple<ll, ll, int>> a(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;
}

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

ll arrival_time(ll Y) {
    VL cur = T;
    cur.pb(Y);
    replr(i, 1, m-1) {
        cur = step(cur, i);
        /*for (ll x : cur) cout << x << " "; cout << endl;*/
    }
    return cur.back();
}

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