結果

問題 No.3072 Speedrun Query
ユーザー 👑 binap
提出日時 2025-03-21 21:51:47
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,172 ms / 2,500 ms
コード長 4,812 bytes
コンパイル時間 4,544 ms
コンパイル使用メモリ 273,836 KB
実行使用メモリ 88,372 KB
最終ジャッジ日時 2025-03-21 21:52:13
合計ジャッジ時間 21,929 ms
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1
other AC * 21
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 #
プレゼンテーションモードにする

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : "");
    return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n
    - 1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
const long long INF = 1001001001001001;
using S = long long;
S _INF(INF);
S _ZERO(0LL);
using F = long long;
S apply(F f, S x){
return f + x;
}
template<typename S, typename F>
struct Dijkstra{
struct Edge{
int from, to;
F cost;
Edge(int from, int to, F cost) : from(from), to(to), cost(cost) {};
};
int n, m;
vector<bool> initialized;
vector<Edge> E;
vector<vector<int>> G;
map<int, vector<S>> dist;
map<int, vector<int>> idx;
Dijkstra(int _n) : n(_n), m(0), initialized(n, false), G(n){}
void add_edge(int from, int to, F cost){
Edge e(from, to, cost);
E.push_back(e);
G[from].emplace_back(m);
m++;
}
void calc(int s){
initialized[s] = true;
dist[s] = vector<S>(n, _INF);
idx[s] = vector<int>(n, -1);
priority_queue<tuple<S, int, int>, vector<tuple<S, int, int>>, greater<tuple<S, int, int>>> pq;
pq.emplace(_ZERO, s, -1);
while(pq.size()){
auto [dist_from, from, index] = pq.top(); pq.pop();
if(dist[s][from] <= dist_from) continue;
dist[s][from] = dist_from;
idx[s][from] = index;
for(int index : G[from]){
int to = E[index].to;
S dist_to = apply(E[index].cost, dist_from);
if(dist[s][to] <= dist_to) continue;
pq.emplace(dist_to, to, index);
}
}
}
int farthest(int s){
if(!initialized[s]) calc(s);
int idx = 0;
rep(i, n) if(dist[s][i] > dist[s][idx]) idx = i;
return idx;
}
S get_dist(int s, int t){
if(!initialized[s]) calc(s);
return dist[s][t];
}
vector<int> restore(int s, int t){
if(!initialized[s]) calc(s);
if(dist[s][t] == _INF) return vector<int>(0);
vector<int> res;
while(idx[s][t] != -1){
auto e = E[idx[s][t]];
res.push_back(idx[s][t]);
t = e.from;
}
reverse(res.begin(), res.end());
return res;
}
};
int main(){
int n, ka, kb;
cin >> n >> ka >> kb;
vector<int> a(ka);
vector<int> b(kb);
cin >> a >> b;
rep(i, ka) a[i]--;
rep(i, kb) b[i]--;
Dijkstra<long long, long long> graph(n + 2);
rep(i, n - 1) graph.add_edge(i, i + 1, 1LL);
rep(i, n - 1) graph.add_edge(i + 1, i, 1LL);
rep(i, ka) graph.add_edge(a[i], n, 0LL);
rep(i, ka) graph.add_edge(n, a[i], 0LL);
rep(i, kb) graph.add_edge(b[i], n + 1, 0LL);
rep(i, kb) graph.add_edge(n + 1, b[i], 0LL);
int q;
cin >> q;
rep(_, q){
int s, t;
cin >> s >> t;
s--; t--;
long long ans = INF;
chmin(ans, (long long)abs(s - t));
chmin(ans, graph.get_dist(n, s) + graph.get_dist(n, t));
chmin(ans, graph.get_dist(n + 1, s) + graph.get_dist(n + 1, t));
chmin(ans, graph.get_dist(n, s) + graph.get_dist(n, n + 1) + graph.get_dist(n + 1, t));
chmin(ans, graph.get_dist(n, t) + graph.get_dist(n, n + 1) + graph.get_dist(n + 1, s));
cout << ans << "\n";
}
return 0;
}
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