제출 #417067

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
417067		yuto1115	Shortcut (IOI16_shortcut)	C++17	71 / 100	2059 ms	1704 KiB

shortcut

#include "shortcut.h"
#include <bits/stdc++.h>
#define rep(i, n) for(ll i = 0; i < ll(n); i++)
#define rep2(i, s, n) for(ll i = ll(s); i < ll(n); i++)
#define rrep(i, n) for(ll i = ll(n)-1; i >= 0; i--)
#define pb push_back
#define eb emplace_back
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(),a.rend()
using namespace std;
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;
using vp = vector<P>;
using vvp = vector<vp>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vs = vector<string>;

template<class T>
bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmax(T &a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

const int inf = 1001001001;
const ll linf = 1001001001001001001;

ll find_shortcut(int n, vi l, vi d, int c) {
    vl L(n);
    rep(i, n - 1) L[i + 1] = L[i] + l[i];
    ll ok = linf, ng = 0;
    auto f = [&](ll x) -> bool {
        bool first = true;
        ll p, q, r, s;
        rep(a, n) rep2(b, a + 1, n) {
                ll nx = x - d[a] - d[b];
                if (L[b] - L[a] <= nx) continue;
                if (c > nx) return false;
                ll w = nx - c;
                ll np = L[a] + L[b] - w;
                ll nq = L[a] + L[b] + w;
                ll nr = -L[a] + L[b] - w;
                ll ns = -L[a] + L[b] + w;
                if (first) {
                    p = np;
                    q = nq;
                    r = nr;
                    s = ns;
                    first = false;
                } else {
                    chmax(p, np);
                    chmin(q, nq);
                    chmax(r, nr);
                    chmin(s, ns);
                }
            }
        if (first) return true;
        rep(i, n) rep2(j, i + 1, n) {
                ll nx = L[i] + L[j];
                ll ny = -L[i] + L[j];
                if (p <= nx and nx <= q and r <= ny and ny <= s) return true;
            }
        return false;
    };
    while (ok - ng > 1) {
        ll mid = (ok + ng) / 2;
        (f(mid) ? ok : ng) = mid;
    }
    return ok;
}

• 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