제출 #1236002

#제출 시각아이디문제언어결과실행 시간메모리
1236002dssfsuper2Magic Tree (CEOI19_magictree)C++20
9 / 100
2096 ms10056 KiB
#include <bits/stdc++.h> using namespace std; #define int long long using pii = pair<int, int>; #define all(x) (x).begin(), (x).end() vector<int> values; vector<int> days; vector<vector<int>> adj; int dpf(int node, int limd){ int res = (days[node]<=limd? values[node] : 0); int res2= 0; int juice=values[node]; int day=days[node]; //chaque truc appele avec 2 valeurs for(auto thing:adj[node]){ if(res!=0)res+=dpf(thing, min(days[node], limd)); res2+=dpf(thing, min((int)1e9, limd)); } return max(res, res2); } void solve(){ int n, m, k;cin>>n>>m>>k; values.assign(n+1, 0); days.assign(n+1, 0); adj.resize(n+1); for(int i = 2;i<=n;i++){ int p;cin>>p; adj[p].push_back(i); } for(int i = 0;i<m;i++){ int v, d, w;cin>>v>>d>>w; values[v]=w; days[v]=d; } cout << dpf(1, (int)1e9) << '\n'; //with sets its total good time merging, but I also need to mantain //like a double log n full I do have it but its so shitty i mean //39 minutes got all ideas exceot one //pretty sure 83 very doable because OOH WAIT //so returns a set<int> //when small to large merging then I can also detect needed day //all values of v distinct thus no vertex >1 fruit that's good //so for each subtree value, days not important UNLESS two on same day then if I want to cut //so cutting something annulates everything cut strictly after //so I will redo dfs, check values after //if values after contain bad bad bad >me juice, do nothing else cut. so //dpf(node) should return maximum juice value I can get after day x //so I should start at the root //then for each fruit I see (basically at every node) //I do:SMALL TO LARGE MERGING //like I will have times // times are very not good //so because those are sets I will swap em and redo //if done correctly with references then a good time } signed main(){ ios_base::sync_with_stdio(false);cin.tie(0); int t = 1; while (t--) { solve(); } }
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