제출 #1125176

#제출 시각아이디문제언어결과실행 시간메모리
1125176AverageAmogusEnjoyerRegions (IOI09_regions)C++17
100 / 100
4962 ms77952 KiB
#pragma GCC optimize("O3,unroll-loops") #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") #include <bits/stdc++.h> using namespace std; using ll = long long; template<class T> bool cmin(T &i, T j) { return i > j ? i=j,true:false; } template<class T> bool cmax(T &i, T j) { return i < j ? i=j,true:false; } mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count()); uniform_int_distribution<int> ui(0, 1 << 30); int rng() { return ui(mrand); } const int N=200001; const int B=50; int pref[B][N]; ll res[B][N]; int order[N]; vector<int> nodes[N]; vector<int> adj[N]; int n,R,q; int regions[N]; int timer=0,tin[N],tout[N]; void dfs(int u=0) { tin[u]=timer++; nodes[regions[u]].push_back(u); for (int &v: adj[u]) { dfs(v); } tout[u]=timer-1; } int dp[N]; void dfs2(int u=0) { for (int &v: adj[u]) { dp[v]+=dp[u]; dfs2(v); } } namespace atcoder { template <class T> struct fenwick_tree { public: fenwick_tree() : _n(0) {} fenwick_tree(int n) : _n(n), data(n) {} void add(int p, T x) { assert(0 <= p && p < _n); p++; while (p <= _n) { data[p - 1] += x; p += p & -p; } } T sum(int l, int r) { assert(0 <= l && l <= r && r <= _n); return sum(r) - sum(l); } private: int _n; vector<T> data; T sum(int r) { T s = 0; while (r > 0) { s += data[r - 1]; r -= r & -r; } return s; } }; } // namespace atcoder using namespace atcoder; fenwick_tree<int> ft(N); int main() { cin >> n >> R >> q; cin >> regions[0]; regions[0]--; for (int i=1;i<n;i++) { int p; cin >> p; adj[--p].push_back(i); cin >> regions[i]; --regions[i]; } dfs(); iota(order,order+R,0); sort(order,order+R,[&](int r1,int r2){ return nodes[r1].size()>nodes[r2].size(); }); for (int i=0;i<min(R,B);i++) { int r=order[i]; for (int &u: nodes[r]) pref[i][tin[u]]++; for (int j=1;j<n;j++) pref[i][j]+=pref[i][j-1]; for (int &u: nodes[r]) dp[u]++; dfs2(); for (int j=0;j<n;j++) if (regions[j]!=r) { res[i][regions[j]]+=dp[j]; } for (int j=0;j<n;j++) dp[j]=0; } auto solve = [&](int r1,int r2) { for (int i=0;i<B;i++) if (order[i]==r1) { return res[i][r2]; } for (int i=0;i<B;i++) if (order[i]==r2) { ll ans=0; for (int &u: nodes[r1]) { ans+=pref[i][tout[u]]-(tin[u]?pref[i][tin[u]-1]:0); } return ans; } for (int &v: nodes[r2]) ft.add(tin[v],1); ll ans=0; for (int &u: nodes[r1]) ans+=ft.sum(tin[u],tout[u]+1); for (int &v: nodes[r2]) ft.add(tin[v],-1); return ans; }; for (int r1,r2;q--;) { cin >> r1 >> r2; --r1,--r2; cout << solve(r1,r2) << endl; // solve naive } }
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