Submission #371544

#	Time	Username	Problem	Language	Result	Execution time	Memory
371544		dolphingarlic	Shortcut (IOI16_shortcut)	C++14	38 / 100	58 ms	748 KiB

shortcut

#include "shortcut.h"

#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

const ll INF = 2e12;

void update(vector<ll> &segtree, int pos, ll val, int node, int l, int r) {
    if (l == r) segtree[node] = val;
    else {
        int mid = (l + r) >> 1;
        if (pos > mid) update(segtree, pos, val, node * 2 + 1, mid + 1, r);
        else update(segtree, pos, val, node * 2, l, mid);
        segtree[node] = max(segtree[node * 2], segtree[node * 2 + 1]);
    }
}

ll query(vector<ll> &segtree, int a, int b, int node, int l, int r) {
    if (l > b || r < a) return -INF;
    if (l >= a && r <= b) return segtree[node];
    int mid = (l + r) >> 1;
    return max(query(segtree, a, b, node * 2, l, mid), query(segtree, a, b, node * 2 + 1, mid + 1, r));
}

ll find_shortcut(int n, vector<int> l, vector<int> d, int c) {
    vector<ll> p(n, 0);
    for (int i = 1; i < n; i++) p[i] = p[i - 1] + l[i - 1];

    // Order the stations by their inequality stuff
    vector<pair<ll, int>> iot(n);
    vector<int> ord(n);
    // Order by d[i] + p[i]
    for (int i = 0; i < n; i++) iot[i] = {d[i] + p[i], i};
    sort(iot.begin(), iot.end());
    for (int i = 0; i < n; i++) ord[iot[i].second] = i;

    // A segtree for each inequality
    vector<ll> seg[2];
    for (int i : {0, 1}) seg[i].resize(4 * n);

    // Binary search for the answer
    ll lo = 0, hi = INF;
    while (lo != hi) {
        ll mid = (lo + hi) >> 1;
        // Reset the segtrees
        for (int i : {0, 1}) fill(seg[i].begin(), seg[i].end(), -INF);

        vector<ll> half_plane(4, -INF);
        for (int i = n - 1; ~i; i--) {
            // Get the index of the first non-ignored station
            int threshold = upper_bound(iot.begin(), iot.end(), make_pair(mid - d[i] + p[i], n)) - iot.begin();
            // Update the half planes
            half_plane[0] = max(half_plane[0], c - mid + d[i] + p[i] + query(seg[0], threshold, n - 1, 1, 0, n - 1));
            half_plane[1] = max(half_plane[1], c - mid + d[i] - p[i] + query(seg[0], threshold, n - 1, 1, 0, n - 1));
            half_plane[2] = max(half_plane[2], c - mid + d[i] + p[i] + query(seg[1], threshold, n - 1, 1, 0, n - 1));
            half_plane[3] = max(half_plane[3], c - mid + d[i] - p[i] + query(seg[1], threshold, n - 1, 1, 0, n - 1));
            // Then add station i to the segtrees
            update(seg[0], ord[i], d[i] + p[i], 1, 0, n - 1);
            update(seg[1], ord[i], d[i] - p[i], 1, 0, n - 1);
        }

        // Check that the half-planes have a common intersection
        bool good = false;
        // Point (x, y) is below the half-plane intersection if y < max(-x + half_plane[0], x + half_plane[1])
        // Point (x, y) is above the half-plane intersection if y > min(-x - half_plane[3], x - half_plane[2])
        for (int i = 0, lower = n - 1, upper = 0; i < n; i++) {
            while (~lower && p[lower] >= max(-p[i] + half_plane[0], p[i] + half_plane[1]))
                lower--;
            while (lower + 1 < n && p[lower + 1] < max(-p[i] + half_plane[0], p[i] + half_plane[1]))
                lower++;
            while (upper < n && p[upper] <= min(-p[i] - half_plane[3], p[i] - half_plane[2]))
                upper++;
            while (~(upper - 1) && p[upper - 1] > min(-p[i] - half_plane[3], p[i] - half_plane[2]))
                upper--;

            if (upper - lower > 1) {
                good = true;
                break;
            }
        }
        if (good) hi = mid;
        else lo = mid + 1;
    }
    return lo;
}

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