제출 #479782

#제출 시각아이디문제언어결과실행 시간메모리
479782jli12345Harbingers (CEOI09_harbingers)C++14
0 / 100
114 ms21336 KiB
#include <bits/stdc++.h> using namespace std; void fastIO(){ ios_base::sync_with_stdio(false); cin.tie(0); } typedef long long ll; #define f first #define s second int N; vector< pair<int, ll> > edges[101000]; pair<ll, ll> city[100100]; // w, t ll dist[100100], dp[100100]; int convex[100100]; int pos = 0; ll m(int ind){ return -dist[ind]; } ll b(int ind){ return dp[ind]; } ll val(int j, int i){ return m(j)*city[i].s+b(j); } ll add(int ind){ return dist[ind]*city[ind].s+city[ind].f; } ll best(int node){ if (node == 1) return 0; int l = 1, r = pos; while (l != r){ int mid = (l+r+1)/2; if (val(convex[mid], node) >= val(convex[mid-1], node)){ r = mid-1; } else { l = mid; } } /* if (node == 3) cout << "best:" << l << " " << val(convex[l], node) << " " << add(node) << "\n"; */ return val(convex[l], node)+add(node); } long double intersect(int aa, int bb){ return (long double)(b(aa)-b(bb))/(m(bb)-m(aa)); } pair<int, ll> insert(int node){ //returns the index of insertion and previous value at that insertion if (node == 1){ convex[1] = 1; pos = 1; return {1, 0}; } int l = 1, r = pos; while (l != r){ int mid = (l+r+1)/2; if (intersect(node, mid) >= intersect(mid, mid-1)){ l = mid; } else { r = mid-1; } } ll pre = convex[l+1]; convex[l+1] = node; pos = l+1; return {l+1, pre}; } void dfs(int node, int pa = 0){ dp[node] = best(node); int prevlen = pos; pair<int, ll> pref = insert(node); /* if (node == 2){ cout << pos << ": "; for (int i = 1; i <= pos; i++){ cout << convex[i] << " "; } cout << "\n"; }*/ for (pair<int, int> nx : edges[node]){ if (nx.f == pa) continue; dist[nx.f] = dist[node]+nx.s; dfs(nx.f, node); } convex[pref.f] = pref.s; pos = prevlen; } int main(){ fastIO(); cin >> N; for (int i = 1; i < N; i++){ int a, b, c; cin >> a >> b >> c; edges[a].push_back({b, c}); edges[b].push_back({a, c}); } for (int i = 1; i < N; i++){ cin >> city[i+1].f >> city[i+1].s; } dfs(1, 0); for (int i = 2; i <= N; i++) cout << dp[i] << " "; cout << "\n"; }
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