제출 #589043

#제출 시각아이디문제언어결과실행 시간메모리
589043radalEvent Hopping (BOI22_events)C++17
10 / 100
313 ms35460 KiB
#include <bits/stdc++.h> //#pragma GCC target("sse,sse2,sse4,avx2") //#pragma GCC optimize("unroll-loops,O2") #define rep(i,l,r) for (int i = l; i < r; i++) #define repr(i,r,l) for (int i = r; i >= l; i--) #define X first #define Y second #define all(x) (x).begin() , (x).end() #define pb push_back #define endl '\n' #define debug(x) cerr << #x << " : " << x << endl; using namespace std; typedef long long ll; typedef long double ld; typedef pair<int,int> pll; constexpr int N = 1e5+10,mod = 1e9+7,inf = 1e9+10,sq = 90; inline int mkay(int a,int b){ if (a+b >= mod) return a+b-mod; if (a+b < 0) return a+b+mod; return a+b; } inline int poww(int a,int k){ if (k < 0) return 0; int z = 1; while (k){ if (k&1) z = 1ll*z*a%mod; a = 1ll*a*a%mod; k >>= 1; } return z; } int nxt[N][20]; pll a[N]; int par[N],d[N]; vector<int> adj[N]; map<int,vector<int>> st,en; void bfs(int v){ par[v] = v; d[v] = 0; queue<int> q; q.push(v); while (!q.empty()){ int u = q.front(); q.pop(); for (int w : adj[u]){ if (par[w] == -1){ par[w] = v; d[w] = d[u]+1; q.push(w); } } } } int main(){ ios :: sync_with_stdio(0); cin.tie(0); int n,q; cin >> n >> q; vector<int> ve; rep(i,1,n+1){ cin >> a[i].X >> a[i].Y; ve.pb(a[i].X); ve.pb(a[i].Y); st[a[i].X].pb(i); en[a[i].Y].pb(i); } set<int> cur; sort(all(ve)); int perv = -1; for (int u : ve){ if (u == perv) continue; perv = u; for (int i : st[u]) cur.insert(i); for (int i : en[u]){ for (int j : cur){ adj[i].pb(j); } } for (int i : en[u]) cur.erase(i); } if (n <= 1000 && q <= 100){ rep(i,0,q){ int s,e; cin >> s >> e; if (s == e){ cout << 0 << endl; continue; } if (a[s].Y == a[e].Y){ cout << 1 << endl; continue; } rep(i,1,n+1) par[i] = -1; bfs(s); if (par[e] == -1) cout << "impossible" << endl; else cout << d[e] << endl; continue; } } else{ memset(par,-1,sizeof par); rep(i,1,n+1) if (par[i] == -1) bfs(i); while (q--){ int s,e; cin >> s >> e; if (s == e){ cout << 0 << endl; continue; } if (a[s].Y == a[e].Y){ cout << 1 << endl; continue; } if (par[s] != par[e]) cout << "impossible" << endl; else cout << d[e]-d[s] << endl; } } return 0; }
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