제출 #558778

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
558778		elazarkoren	Shortcut (IOI16_shortcut)	C++17	0 / 100	1 ms	340 KiB

shortcut

#include "shortcut.h"
#include <bits/stdc++.h>
#define x first
#define y second
#define all(v) v.begin(), v.end()
#define chkmin(a, b) a = min(a, b)
#define chkmax(a, b) a = max(a, b)
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> pii;
typedef vector<pii> vii;

const int MAX_N = 1e6 + 5;
const ll infinity = 1e18;

ll prefix[MAX_N];

inline ll Dist(int l, int r) {
    if (l > r) swap(l, r);
    return prefix[r] - prefix[l];
}

ll farl[MAX_N], farr[MAX_N];
ll dpl[MAX_N], dpr[MAX_N];

ll find_shortcut(int n, vi l, vi d, int c2) {
    for (int i = 1; i < n; i++) prefix[i] = prefix[i - 1] + l[i - 1];
    farl[0] = dpl[0] = d[0];
    for (int i = 1; i < n; i++) {
        farl[i] = max<ll>(farl[i - 1] + l[i - 1], d[i]);
        dpl[i] = max<ll>(dpl[i - 1], farl[i - 1] + l[i - 1] + d[i]);
    }
    dpr[n - 1] = farr[n - 1] = d[n - 1];
    for (int i = n - 2; i >= 0; i--) {
        farr[i] = max<ll>(farr[i + 1] + l[i], d[i]);
        dpr[i] = max<ll>(dpr[i + 1], farr[i + 1] + l[i] + d[i]);
    }
    ll ans = infinity;
    for (int i = 0; i < n; i++) {
        //vi s = {0};
        ll circle = c2;
        for (int j = i + 1; j < n; j++) {
            //ll far_left = farl[i], far_right = farr[j];
            ll c = min<ll>(c2, Dist(i, j));
            ll curr = max({dpl[i], dpr[j], farl[i] + farr[j] + c});
            for (int k = i; k <= j; k++) {
               ll x = min(Dist(i, k), Dist(k, j) + c) + d[k];
               chkmax(curr, x + farl[i]);
               x = min(Dist(k, j), Dist(i, k) + c) + d[k];
               chkmax(curr, x + farr[j]);
            }
            //s.push_back(s.back() + l[j - 1]);
            //s.push_back(s.back() + c);
            priority_queue<pii> p1, p2, p3;
            for (int k = i + 1; k < j; k++) {
                p3.push({prefix[k] + d[k], k});
            }
            p2.push({-prefix[i] + d[i], i});
            circle += l[j - 1];
            //ll circle = s.back();
            //int sz = s.size();
            for (int k = i + 1, h = i; k <= j; k++) {
                while (h < k && prefix[k] - prefix[h] > circle / 2) {
                    p1.push({prefix[h] + d[h], h});
                    h++;
                }
                while (!p2.empty() && p2.top().y < h) p2.pop();
                while (!p3.empty() && p3.top().y < k) p3.pop();
                if (!p1.empty()) {
                    chkmax(curr, circle - prefix[k] + p1.top().x + d[k]);
                }
                if (!p2.empty()) {
                    chkmax(curr, prefix[k] + p2.top().x + d[k]);
                }
                if (!p3.empty()) {
                    chkmax(curr, p3.top().x - prefix[k] + d[k]);
                }
                p2.push({-prefix[k] + d[k], k});
                //chkmin(diameter, circle - (s[i] - s[h + 1]));
            }
            //s.pop_back();
            //chkmax(curr, diameter);
//            ll curr = 0;
//            for (int a = 0; a < n; a++) {
//                for (int b = a + 1; b < n; b++) {
//                    ll x = min({Dist(a, b), min(Dist(a, i), Dist(a, j)) + min(Dist(b, i), Dist(b, j)) + c});
//                    chkmax(curr, x + d[a] + d[b]);
//                }
//            }
            chkmin(ans, curr);
        }
    }
    return ans;
}

• 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