Submission #1010242

#TimeUsernameProblemLanguageResultExecution timeMemory
1010242GrindMachine도로 폐쇄 (APIO21_roads)C++17
0 / 100
2094 ms16852 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long int ll; typedef long double ld; typedef pair<int,int> pii; typedef pair<ll,ll> pll; #define fastio ios_base::sync_with_stdio(false); cin.tie(NULL) #define pb push_back #define endl '\n' #define sz(a) (int)a.size() #define setbits(x) __builtin_popcountll(x) #define ff first #define ss second #define conts continue #define ceil2(x,y) ((x+y-1)/(y)) #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define yes cout << "Yes" << endl #define no cout << "No" << endl #define rep(i,n) for(int i = 0; i < n; ++i) #define rep1(i,n) for(int i = 1; i <= n; ++i) #define rev(i,s,e) for(int i = s; i >= e; --i) #define trav(i,a) for(auto &i : a) template<typename T> void amin(T &a, T b) { a = min(a,b); } template<typename T> void amax(T &a, T b) { a = max(a,b); } #ifdef LOCAL #include "debug.h" #else #define debug(...) 42 #endif /* read some solutions a long time ago, dont remember much from there (remember seeing sqrt complexity in some solutions, but im not sure) */ const int MOD = 1e9 + 7; const int N = 1e5 + 5; const int inf1 = int(1e9) + 5; const ll inf2 = ll(1e18) + 5; #include "roads.h" vector<pll> adj[N]; ll dp[N][2]; vector<ll> order; void dfs1(ll u){ for(auto [v,w] : adj[u]){ pll px = {u,w}; adj[v].erase(find(all(adj[v]),px)); dfs1(v); } order.pb(u); } vector<ll> minimum_closure_costs(int n, vector<int> U, vector<int> V, vector<int> W){ rep(i,n-1){ int u = U[i], v = V[i], w = W[i]; adj[u].pb({v,w}), adj[v].pb({u,w}); } dfs1(1); vector<ll> ans; rep(k,n){ trav(u,order){ dp[u][0] = dp[u][1] = 0; vector<ll> vals; ll sum = 0; for(auto [v,w] : adj[u]){ sum += w+dp[v][1]; vals.pb(-w-dp[v][1]+dp[v][0]); } sort(all(vals)); rep(i,k-1){ if(i < sz(vals)){ sum += min(vals[i],0ll); } else{ break; } } dp[u][0] = dp[u][1] = sum; if(k-1 < sz(vals) and k-1 >= 0){ dp[u][1] += min(vals[k-1],0ll); } } ans.pb(min(dp[1][0],dp[1][1])); } ll change = 0; rep(i,n-1) change += (ans[i] != ans[i+1]); assert(change <= 100); return ans; }
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