Submission #225390

#	Time	Username	Problem	Language	Result	Execution time	Memory
225390		VEGAnn	Shortcut (IOI16_shortcut)	C++14	71 / 100	496 ms	2432 KiB

shortcut

#include <bits/stdc++.h>
#include "shortcut.h"
using namespace std;
typedef long long ll;
const int N = 3010;
const ll OO = 1e18;
ll pf[N], dst[N], c;
int n, mem_l[N], mem_r[N];

bool ok(ll mx){
    bool was = 0;
    ll u = -1, d = -1, l = -1, r = -1;

    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if (pf[j] - pf[i] + dst[j] + dst[i] > mx){
                ll k = mx - dst[i] - dst[j] - c;
                if (k < 0) return 0;

                ll up = pf[j] + k, down = pf[j] - k;

                ll cu = up + pf[i];
                ll cd = down + pf[i];
                ll cl = pf[i] - (up + pf[i]) / 2;
                ll cr = pf[i] - (down + pf[i]) / 2;

                if (!was){
                    was = 1;
                    u = cu; d = cd; l = cl; r = cr;
                } else {
                    u = min(u, cu);
                    d = max(d, cd);
                    if (u < d) return 0;
                    l = max(l, cl);
                    r = min(r, cr);
                    if (r < l) return 0;
                }
            }

    if (!was) return 1;

    int l1 = 0, r1 = -1;
    int l2 = -1, r2 = 0;

    for (int i = n - 1; i >= 0; i--){
        while (l1 < n && pf[l1] + pf[i] < d)
            l1++;
        while (r1 + 1 < n && pf[r1 + 1] + pf[i] <= u)
            r1++;

        mem_l[i] = l1;
        mem_r[i] = r1;
    }

    for (int i = 0; i < n; i++) {
        while (l2 < n - 1 && (pf[i] - pf[l2 + 1]) / 2 >= l)
            l2++;
        while (r2 < n && (pf[i] - pf[r2]) / 2 > r)
            r2++;

        int j1 = max(mem_l[i], r2), j2 = min(mem_r[i], l2);

        if (j1 > j2) continue;

        if (j1 == j2){
            if (j1 != i)
                return 1;
        } else return 1;
    }

    return 0;
}

long long find_shortcut(int N, std::vector<int> l, std::vector<int> d, int C){
    n = N;
    c = C * 2;

    assert(n <= 3000);

    pf[0] = 0;

    for (int i = 0; i < n - 1; i++) {
        pf[i + 1] = pf[i] + l[i] * 2;
        dst[i] = d[i] * 2;
    }

    dst[n - 1] = d[n - 1] * 2;

    // smth smaller
    ll l1 = 0, r1 = OO;

    while (l1 < r1){
        ll md = (l1 + r1) >> 1;

        if (ok(md))
            r1 = md;
        else l1 = md + 1;
    }

    ok(22);

    return l1 / 2;
}

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