제출 #955194

#제출 시각아이디문제언어결과실행 시간메모리
955194amine_arouaRailway Trip 2 (JOI22_ho_t4)C++17
0 / 100
2043 ms11092 KiB
#include <bits/stdc++.h> //#pragma GCC optimize("O3") //#pragma GCC optimize("unroll-loops") using namespace std; #define int long long #define vi vector<int> #define vl vector<long long> #define vii vector<pair<int,int>> #define vll vector<pair<long long,long long>> #define pb push_back #define ll long long #define ld long double #define nl '\n' #define boost ios::sync_with_stdio(false) #define mp make_pair #define se second #define fi first #define fore(i, y) for(int i = 0; i < y; i++) #define forr(i,x,y) for(int i = x;i<=y;i++) #define forn(i,y,x) for(int i = y; i >= x; i--) #define all(v) v.begin(),v.end() #define sz(v) (int)v.size() #define clr(v,k) memset(v,k,sizeof(v)) #define rall(v) v.rbegin() , v.rend() #define pii pair<int,int> #define pll pair<ll , ll> const ll MOD = 1e9 + 7; const ll INF = 1e18 + 1; ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (gcd) ll lcm(ll a , ll b) {return a * (b / gcd(a , b));} // least common multiple (lcm) // HERE IS THE SOLUTION vi mxTree , mnTree; int n; void build() { while(__builtin_popcount(n) != 1) n++; mxTree.assign(2*n , 0); mnTree.assign(2*n , n + 1); } void upd(int node , int L , int R , int l , int r , int x , bool type) { if(l <= L && R <= r) { if(!type) mxTree[node] = max(mxTree[node] , x); else mnTree[node] = min(mnTree[node] , x); return ; } if(l > R || L > r) return ; int mid = (L + R)/2; upd(2*node , L , mid , l , r , x , type); upd(2*node + 1 , mid + 1 , R , l , r , x , type); } int getMax(int i) { int ans = i; for(int j = (i + n); j>= 1;j/=2) { ans = max(ans , mxTree[j]); } return ans; } int getMin(int i) { int ans = i; for(int j = (i + n); j>= 1;j/=2) { ans = min(ans , mnTree[j]); } return ans; } signed main() { boost; cin.tie(0); cout.tie(0); int N , K , M; cin>>N>>K>>M; n = N +1; build(); vector<vi> adj(N + 1); forr(i , 1 , N) { adj[i] = {i , i}; } fore(i , M) { int A , B; cin>>A>>B; if(A < B) { upd(1 , 0 , n - 1 , A , min(B - 1 , A + K - 1) , B , 0); } else { upd(1 , 0 , n - 1 , max(B + 1 , A - K + 1) , A , B , 1); } } forr(i , 1 , N) { adj[i][0] = getMin(i); adj[i][1] = getMax(i); } int Q; cin>>Q; while(Q--) { int S , T; cin>>S>>T; queue<int> q; q.push(S); vi dist(N + 1 , INF); vector<bool> vis(N + 1 , 0); dist[S] = 0; while(!q.empty()) { int cur = q.front(); q.pop(); if(vis[cur]) continue; vis[cur] = 1; if(cur == T) break; for(int u = adj[cur][0] ; u <= adj[cur][1]; u++) { if(dist[u] > dist[cur] + 1) dist[u] = dist[cur] + 1; } if(adj[cur][1] < N) q.push(adj[cur][1] + 1); if(adj[cur][0] > 1) q.push(adj[cur][0] - 1); } cout<<(dist[T] == INF ? -1 : dist[T])<<nl; } }
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