結果

問題 No.2242 Cities and Teleporters
ユーザー dyktr_06dyktr_06
提出日時 2023-03-10 22:31:09
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 2,509 ms / 3,000 ms
コード長 5,763 bytes
コンパイル時間 4,869 ms
コンパイル使用メモリ 275,416 KB
実行使用メモリ 28,404 KB
最終ジャッジ日時 2024-09-18 04:41:38
合計ジャッジ時間 24,820 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
5,248 KB
testcase_01 AC 2 ms
5,376 KB
testcase_02 AC 2 ms
5,376 KB
testcase_03 AC 2 ms
5,376 KB
testcase_04 AC 2 ms
5,376 KB
testcase_05 AC 1,800 ms
28,292 KB
testcase_06 AC 1,034 ms
28,404 KB
testcase_07 AC 1,664 ms
28,184 KB
testcase_08 AC 2,509 ms
28,200 KB
testcase_09 AC 2,435 ms
28,160 KB
testcase_10 AC 304 ms
28,188 KB
testcase_11 AC 250 ms
28,248 KB
testcase_12 AC 249 ms
28,160 KB
testcase_13 AC 476 ms
28,180 KB
testcase_14 AC 819 ms
28,284 KB
testcase_15 AC 445 ms
28,160 KB
testcase_16 AC 508 ms
28,204 KB
testcase_17 AC 1,211 ms
28,264 KB
testcase_18 AC 330 ms
27,888 KB
testcase_19 AC 393 ms
27,748 KB
testcase_20 AC 414 ms
27,144 KB
testcase_21 AC 305 ms
27,204 KB
testcase_22 AC 320 ms
27,200 KB
testcase_23 AC 347 ms
28,288 KB
testcase_24 AC 337 ms
28,164 KB
testcase_25 AC 334 ms
28,288 KB
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
#include <atcoder/all>

using namespace std;
using namespace atcoder;

#define overload4(_1, _2, _3, _4, name, ...) name
#define rep1(n) for(int i = 0; i < (int)(n); ++i)
#define rep2(i, n) for(int i = 0; i < (int)(n); ++i)
#define rep3(i, a, b) for(int i = (a); i < (int)(b); ++i)
#define rep4(i, a, b, c) for(int i = (a); i < (int)(b); i += (c))
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)

#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()
#define Sort(a) sort(a.begin(), a.end())
#define RSort(a) sort(a.rbegin(), a.rend())

typedef long long int ll;
typedef unsigned long long ul;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector<char> vc;
typedef vector<string> vst;
typedef vector<double> vd;
typedef vector<long double> vld;
typedef pair<long long, long long> P;

template<class T> long long sum(const T& a){ return accumulate(a.begin(), a.end(), 0LL); }
template<class T> auto min(const T& a){ return *min_element(a.begin(), a.end()); }
template<class T> auto max(const T& a){ return *max_element(a.begin(), a.end()); }

const long long MINF = 0x7fffffffffff;
const long long INF = 0x1fffffffffffffff;
const long long MOD = 998244353;
const long double EPS = 1e-9;
const long double PI = acos(-1);
 
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p){ is >> p.first >> p.second; return is; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){ os << "(" << p.first << ", " << p.second << ")"; return os; }
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for(T &in : v) is >> in; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v){ for(int i = 0; i < (int) v.size(); ++i){ os << v[i] << (i + 1 != (int) v.size() ? " " : ""); } return os; }
template <typename T, typename S> ostream &operator<<(ostream &os, const map<T, S> &mp){ for(auto &[key, val] : mp){ os << key << ":" << val << " "; } return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &st){ auto itr = st.begin(); for(int i = 0; i < (int)st.size(); ++i){ os << *itr << (i + 1 != (int)st.size() ? " " : ""); itr++; } return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &st){ auto itr = st.begin(); for(int i = 0; i < (int)st.size(); ++i){ os << *itr << (i + 1 != (int)st.size() ? " " : ""); itr++; } return os; }
template <typename T> ostream &operator<<(ostream &os, queue<T> q){ while(q.size()){ os << q.front() << " "; q.pop(); } return os; }
template <typename T> ostream &operator<<(ostream &os, deque<T> q){ while(q.size()){ os << q.front() << " "; q.pop_front(); } return os; }
template <typename T> ostream &operator<<(ostream &os, stack<T> st){ while(st.size()){ os << st.top() << " "; st.pop(); } return os; }
template <class T, class Container, class Compare> ostream &operator<<(ostream &os, priority_queue<T, Container, Compare> pq){ while(pq.size()){ os << pq.top() << " "; pq.pop(); } return os; }

template<class T, class U> inline T vin(T& vec, U n) { vec.resize(n); for(int i = 0; i < (int) n; ++i) cin >> vec[i]; return vec; }
template<class T> inline void vout(T vec, string s = "\n"){ for(auto x : vec) cout << x << s; }
template<class... T> void in(T&... a){ (cin >> ... >> a); }
void out(){ cout << '\n'; }
template<class T, class... Ts> void out(const T& a, const Ts&... b){ cout << a; (cout << ... << (cout << ' ', b)); cout << '\n'; }
template<class T, class U> void inGraph(vector<vector<T>>& G, U n, U m, bool directed = false){ G.resize(n); for(int i = 0; i < m; ++i){ int a, b; cin >> a >> b; a--, b--; G[a].push_back(b); if(!directed) G[b].push_back(a); } }

ll n;
vll h, t;

ll d[200001][20];

void input(){
    in(n);
    vin(h, n);
    vin(t, n);
}
 
void solve(){
    vector<tuple<ll, ll, ll>> p(n);
    rep(i, n){
        p[i] = {t[i], h[i], i};
    }
    RSort(p);
    vll mn(n + 1, INF);
    rep(i, n){
        mn[i + 1] = min(mn[i], get<1>(p[i]));
    }
    RSort(p);
    int logK = 1;
    while ((1LL << logK) <= n) logK++;
    logK++;
    // doubling[k][i] : i番目から 2^k 進んだ町
    vector<vector<int>> doubling(logK, vector<int>(n));
    rep(i, n){
        ll ok = n, ng = -1;
        while(abs(ok - ng) > 1){
            ll mid = (ok + ng) / 2;
            if(mn[mid + 1] <= t[i]){
                ok = mid;
            }else{
                ng = mid;
            }
        }
        ll nxt = i;
        if(ok != n){
            if(t[i] < get<0>(p[ok])){
                nxt = get<2>(p[ok]);
            }
        }
        doubling[0][i] = nxt;
    }
    rep(k, logK - 1){
        rep(i, n){
            doubling[k + 1][i] = doubling[k][doubling[k][i]];
        }
    }
    ll q; in(q);
    while(q--){
        ll a, b; in(a, b); a--; b--;
        ll mx = t[doubling[logK - 1][a]];
        if(h[b] > mx){
            out(-1);
            continue;
        }
        ll ok = n, ng = -1;
        while(abs(ok - ng) > 1){
            ll mid = (ok + ng) / 2;
            int now = a, K = mid;
            for(int k = 0; K > 0; k++){
                if (K & 1) now = doubling[k][now];
                K = K >> 1;
            }
            if(h[b] <= t[now]){
                ok = mid;
            }else{
                ng = mid;
            }
        }
        out(ok + 1);
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(20);
    
    input();
    solve();
}
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