Submission #1195757

#	Time	Username	Problem	Language	Result	Execution time	Memory
1195757		madamadam3	Boxes with souvenirs (IOI15_boxes)	C++20	0 / 100	2097 ms	408 KiB

boxes

#include "boxes.h"
#include <bits/stdc++.h>

using namespace std;

#define FOR(i, a, b) for (int i = a; i < b; i++)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define srt(x) sort(all(x))

typedef long long ll;
using vi = vector<int>;
using vl = vector<ll>;

// probably binary search?

ll n, k, l;
vl p;
vl nxt, prev;

ll sim_dist(ll p1, ll p2) {
    ll lmoves = 0, rmoves = 0;
    ll lpos = p1, rpos = p1;

    while (lpos != p2) {
        lpos--;
        if (lpos < 0) lpos = l - 1;
        lmoves++;
    }

    while (rpos != p2) {
        rpos++;
        if (rpos >= l) rpos = 0;
        rmoves++;
    }

    return min(lmoves, rmoves);
}

ll dist(ll p1, ll p2) {
    if (p1 == p2) return 0;
    if (p1 < p2) return dist(p2, p1);
    else return min(p1 - p2, (l - p1) + p2); 
}

// going left for kL teams and coming back
ll sim_left(int kL, deque<int> &cur) {
    ll ansL = 0, pL = 0;
    
    FOR(i, 0, kL) {
        if (pL == 0) ansL += l - cur.back();
        else ansL += pL - cur.back();

        pL = cur.back();
        cur.pop_back();
    }

    ansL += sim_dist(pL, 0);
    return ansL;
}

// going right for kR teams and coming back
ll sim_right(int kR, deque<int> &cur) {
    ll ansR = 0, pR = 0;
    
    FOR(i, 0, kR) {
        ansR += cur.front() - pR;
        pR = cur.front();
        cur.pop_front();
    }
    ansR += sim_dist(pR, 0);

    return ansR;
}

ll delivery(int N, int K, int L, int P[]) {
    n = N, k = K, l = L;

    FOR(i, 0, n) p.pb(P[i]);
    srt(p);

    vl answers;
    FOR(i, 0, n+1) {
        int kL = i, kR = n - i;

        deque<int> cur = deque<int>(all(p));
        answers.pb(sim_left(kL, cur) + sim_right(kR, cur));
    }

    return *min_element(all(answers));

    // return min(*max_element(all(p)) + dist(*max_element(all(p)), 0), l - *min_element(all(p)) + dist(*min_element(all(p)), 0));
    // ll max_pos = *max_element(all(p));
    // ll min_pos = *min_element(all(p));
    // ll min_dist = dist(min_pos, l);

    // return min({l, 2LL * max_pos, 2LL * min_dist});
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30