제출 #372234

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
372234		dolphingarlic	Shortcut (IOI16_shortcut)	C++14	97 / 100	2086 ms	55916 KiB

shortcut

#include "shortcut.h"

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

const ll INF = 1e15;

ll p[1000000], half_plane[4];
pair<ll, int> ord1[1000000], ord2[1000000];

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

    // Order by d[i] + p[i] and d[i] - p[i]
    for (int i = 0; i < n; i++) ord1[i] = {d[i] + p[i], i}, ord2[i] = {d[i] - p[i], i};
    sort(ord1, ord1 + n);
    sort(ord2, ord2 + n);

    // Binary search for the answer
    ll lo = 0, hi = INF;
    while (lo != hi) {
        ll mid = (lo + hi) >> 1;

        half_plane[0] = half_plane[1] = half_plane[2] = half_plane[3] = -INF;
        ll mx1 = -INF, mx2 = -INF, smx1 = -INF, smx2 = -INF;
        for (int i = 0, j = n - 1; i < n; i++) {
            while (~j && ord1[i].first + ord2[j].first > mid) {
                ll tmp1 = d[ord2[j].second] + p[ord2[j].second];
                if (tmp1 > mx1) smx1 = mx1, mx1 = tmp1;
                else if (tmp1 > smx1) smx1 = tmp1;

                ll tmp2 = d[ord2[j].second] - p[ord2[j].second];
                if (tmp2 > mx2) smx2 = mx2, mx2 = tmp2;
                else if (tmp2 > smx2) smx2 = tmp2;

                j--;
            }
            // Update the half planes
            ll use1 = (d[ord1[i].second] + p[ord1[i].second] == mx1 ? smx1 : mx1);
            ll use2 = (d[ord1[i].second] - p[ord1[i].second] == mx2 ? smx2 : mx2);
            half_plane[0] = max(half_plane[0], c - mid + d[ord1[i].second] + p[ord1[i].second] + use1);
            half_plane[1] = max(half_plane[1], c - mid + d[ord1[i].second] - p[ord1[i].second] + use1);
            half_plane[2] = max(half_plane[2], c - mid + d[ord1[i].second] + p[ord1[i].second] + use2);
            half_plane[3] = max(half_plane[3], c - mid + d[ord1[i].second] - p[ord1[i].second] + use2);
        }

        // 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: 33 / 33
• Subtask 6: 22 / 22
• Subtask 7: 4 / 4
• Subtask 8: 0 / 3