제출 #1210069

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1210069		sano	Shortcut (IOI16_shortcut)	C++20	0 / 100	2096 ms	416 KiB

shortcut

#include "shortcut.h"
#include<iostream>
#include<vector>
#include<queue>
#include<deque>
#include<string>
#include<fstream>
#include<algorithm>
#include <iomanip>
#include<map>
#include <set>
#include <unordered_map>
#include <stack>
#include <unordered_set>
#include <cmath>
#include <cstdint>
#include <cassert>
#include <bitset>
#include <random>
#include <chrono>
#include <cstring>
#define shit short int
#define ll long long
#define ld long double
//#define int ll
#define For(i, n) for(int i = 0; i < (int)n; i++)
#define ffor(i, a, n) for(int i = (int)a; i < (int)n; i++)
#define rfor(i, n) for(int i = (int)n; i >= (int)0; i--)
#define rffor(i, a, n) for(int i = (int)n; i >= (int)a; i--)
#define vec vector
#define ff first
#define ss second
#define pb push_back
#define pii pair<ll, ll>
#define pld pair<ld, ld>
#define NEK 2000000000000000
#define mod 1000000007
#define mod2 1000000009
#define rsz resize 
#define prv 43
#define prv2 47
#define D 8
#define trav(a,x) for (auto& a: x)
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define all(x) (x).begin(), (x).end()
#define sig 0.0000001

using namespace std;

class intervalac {
    ll n;
    vec<ll> l, r, in;
public:
    intervalac(ll vel) {
        n = 1;
        while (n < vel) n *= 2;
        l.resize(2 * n); r.resize(2 * n); in.resize(2 * n, (-1) * NEK);
        For(i, n) {
            l[i + n] = r[i + n] = i;
        }
        rffor(i, 1, (n - 1)) {
            l[i] = l[i * 2];
            r[i] = r[i * 2 + 1];
        }
        return;
    }
    void zmen(ll a, ll b) {
        a += n;
        in[a] = b;
        a /= 2;
        while (a) {
            in[a] = max(in[a * 2], in[a * 2 + 1]);
            a /= 2;
        }
        return;
    }
    ll daj(ll a, ll b, ll s = 1) {
        if (l[s] > b || r[s] < a) return (-1) * NEK;
        if (a <= l[s] && r[s] <= b) return in[s];
        return max(daj(a, b, s * 2), daj(a, b, s * 2 + 1));
    }
};

bool check1(vec<ll>&d, vec<ll>&ps, ll k, ll mid, ll x, ll y, ll c) {
    ll n = d.size();
    ll priamo = (ps[mid] - ps[k]);
    ll otocka = (ps[k] - ps[x]) + (ps[y] - ps[mid]) + c;
    return priamo <= otocka;
}

bool check2(vec<ll>& d, vec<ll >& ps, ll k, ll mid, ll x, ll y, ll c) {
    ll n = d.size();
    ll priamo = (ps[k] - ps[mid]);
    ll otocka = (ps[y] - ps[k]) + (ps[mid] - ps[x]) + c;
    return priamo <= otocka;
}

ll find_shortcut_rychle(int n1, vec<int> len, vec<int> d1, int c1) {
    ll n = n1;
    ll c = c1;
    vec<ll> d;
    For(i, d1.size()) d.push_back(d1[i]);
    vec<ll> ps(n);
    ps[0] = 0;
    For(i, (n - 1)) {
        ps[i + 1] = ps[i] + len[i];
    }
    vec<ll> mpref(n), msuf(n);
    ll mojo = d[0] - ps[0];
    mpref[0] = d[0];
    ffor(i, 1, n) {
        mpref[i] = mojo + ps[i] + d[i];
        mpref[i] = max(mpref[i], d[i]);
        mpref[i] = max(mpref[i], mpref[i - 1]);
        mojo = max(mojo, d[i] - ps[i]);
    }
    mojo = ps.back() + d.back();
    msuf[n - 1] = d[n - 1];
    rfor(i, (n-2)) {
        msuf[i] = mojo - ps[i] + d[i];
        msuf[i] = max(msuf[i], d[i]);
        msuf[i] = max(msuf[i], msuf[i + 1]);
        mojo = max(mojo, d[i] + ps[i]);
    }
    intervalac seg1(n);
    intervalac seg2(n);
    For(i, n) {
        seg1.zmen(i, ((-1) * ps[i]) + d[i]);
        seg2.zmen(i, ps[i] + d[i]);
    }
    ll odp = NEK;
    For(i, n) {
        odp = max(odp, d[i]);
    }
    For(x, n) {
        ffor(y, x+1, n) {
            //spajame x a y
            //nalavo od x
            ll lavo = seg1.daj(0, x) + ps[x];
            //napravo od y
            ll pravo = seg2.daj(y, n) - ps[y];
            ll maxi = pravo + lavo + min(ps[y] - ps[x], c);
            maxi = max(maxi, mpref[x]);
            maxi = max(maxi, msuf[y]);
            for (ll k = x+1; k < y; k++) {
                maxi = max(maxi, lavo + min(ps[k] - ps[x], ps[y] - ps[k] + c) + d[k]);
                maxi = max(maxi, pravo + min(ps[y] - ps[k], ps[k] - ps[x] + c) + d[k]);
                //2 situacie -> sme blizsie ku x potom binarne vyhladame najpravejsieho ku ktoremu chceme ist peso
                 if (ps[k] - ps[x] <= (ps[y] - ps[k])) {
                    ll l = k, r = y;
                    while (l < r) {
                        ll mid = (l + r + 1) / 2;
                        if (check1(d, ps, k, mid, x, y, c)) l = mid;
                        else r = mid - 1;
                    }
                    //od x po l chceme ist priamo, inak sa chceme otocit
                    //x az (k-1)
                    if(x <= (k-1)) maxi = max(maxi, seg1.daj(x, k-1) + ps[k] + d[k]);
                    //k+1 az l
                    if((k+1) <= l) maxi = max(maxi, seg2.daj(k+1, l) - ps[k] + d[k]);
                    //l+1 az y
                    if((l+1) <= y) maxi = max(maxi, seg1.daj(l+1, y) + ps[y] + ps[k] - ps[x] + c + d[k]);
                }

                else {
                    //sme blizsie ku y potom binarne vyhladame najlavajsieho ku ktoremu chceme ist peso
                    ll l = x, r = y;
                    while (l < r) {
                        ll mid = (l + r) / 2;
                        if (check2(d, ps, k, mid, x, y, c)) r = mid;
                        else l = mid + 1;
                    }
                    //k+1 az y
                    if((k+1) <= y) maxi = max(maxi, seg2.daj(k+1, y) - ps[k] + d[k]);
                    //l az k-1
                    if(l <= (k-1)) maxi = max(maxi, seg1.daj(l, k-1) + ps[k] + d[k]);
                    //x az l-1
                    if(x <= (l-1)) maxi = max(maxi, seg2.daj(x, l-1) - ps[x] + c + ps[y] - ps[k] + d[k]);
                }
            }
            odp = min(odp, maxi);
        }
    }
    return odp;
}

ll find_shortcut(int n1, vec<int> len1, vec<int>d1, int c1) {
    ll n = n1;
    ll c = c1;
    vec<ll> d;
    vec<ll> len;
    For(i, len1.size()) len.push_back(len1[i]);
    For(i, d1.size()) d.push_back(d1[i]);
    ll odp = NEK;
    For(x, n) {
        ffor(y, x + 1, n) {
            ll max_dist = 0;
            For(k, n) {
                priority_queue<pii> q;
                q.push({ 0, k });
                vec<bool> bol(n, 0);
                while (!q.empty()) {
                    ll som = q.top().ss; ll dist = q.top().ff * (-1); q.pop();
                    if (bol[som]) continue;
                    bol[som] = 1;
                    if(k != som) max_dist = max(max_dist, dist + d[k] + d[som]);
                    if (som != 0) q.push({ (dist + len[som-1]) * (-1), som - 1 });
                    if (som != (n - 1)) q.push({ (dist + len[som]) * (-1), som + 1 });
                    if (som == x) q.push({ (dist + c) * (-1), y });
                    if (som == y) q.push({ (dist + c) * (-1), x });
                }
                
            }
            odp = min(odp, max_dist);
        }
    }
    return odp;
}
/*
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, c; 
    while (1) {
        //cin >> n >> c;
        n = 20;
        c = rand() % 10;
        vec<int> l(n - 1), d(n);
        For(i, (n - 1)) {
            //cin >> l[i];
            l[i] = rand() % 10;
        }
        For(i, n) {
            //cin >> d[i];
            d[i] = rand() % 10;
        }
        ll odp1 = find_shortcut_rychle(n, l, d, c);
        ll odp2 = find_shortcut(n, l, d, c);
        if (odp1 != odp2) {
            cout << odp1 << ' ' << odp2 << '\n';
            cout << n << ' ' << c << '\n';
            For(i, l.size()) cout << l[i] << ' ';
            cout << '\n';
            For(i, d.size()) cout << d[i] << ' ';
            cout << '\n';
            odp1 = find_shortcut_rychle(n, l, d, c);
            odp2 = find_shortcut(n, l, d, c);
        }
        cout << odp2 << '\n';
        //return 0;
    }
    return 0;
}*/

컴파일 시 표준 에러 (stderr) 메시지

shortcut.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~
shortcut_c.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~

• Subtask 1: 0 / 9
• Subtask 2: 0 / 14
• Subtask 3: 0 / 8
• Subtask 4: 0 / 7
• Subtask 5: 0 / 33
• Subtask 6: 0 / 22
• Subtask 7: 0 / 4
• Subtask 8: 0 / 3