Submission #970841

#	Time	Username	Problem	Language	Result	Execution time	Memory
970841		AlperenT_	Road Closures (APIO21_roads)	C++17	31 / 100	2058 ms	53708 KiB

roads

#include "roads.h"
#include <bits/stdc++.h>

#pragma GCC optimize("O3,unroll-loops")
#define pb push_back
#define F first
#define S second 
#define ld long double
#define aint(a) a.begin(),a.end()
#define pii pair <int,int>
#define PII pair<pii , pii>
#define ll long long 
#define sz(v) (int)v.size()
#define rep(i , a , b) for(int i=a;i <= (b);i++)
#define per(i , a , b) for(int i=a;i >= (b);i--)
#define deb(x) cout <<#x << " : " << x << "\n" ;
using namespace std ;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll maxn = 1e6 , maxq = 32, inf = 1e15+10 , lg = 19  ,sq = 707 ,mod = 998244353 ;
vector <pii> G[maxn] ;
ll dp[maxn][2]  , k;

void dfs(int v , int W = 0 , int p = -1){
  dp[v][0] = dp[v][1] = 0;
  int x = max(0ll,sz(G[v])-k) ;
  priority_queue <ll> pq ;
  for(auto [u,w] : G[v]){
    if(u == p)continue ;
    dfs(u , w , v) ;
    if(dp[u][0] <= dp[u][1]){
      dp[v][0] += dp[u][0] ;
      x=max(x-1 , 0) ;
    }else{
      dp[v][0] += dp[u][1] ;
      pq.push((-dp[u][1] + dp[u][0]));
      while(sz(pq) > x)pq.pop();
    }
  }
  dp[v][1] = dp[v][0] ;
  dp[v][0] += W ; 
  if(x==0)return ;
  while(sz(pq) > x)pq.pop();
  if(sz(pq) < x){
    while(sz(pq)){
      dp[v][0] += pq.top() ; pq.pop() ;
    }
    dp[v][1] = inf ;
  }else{
    bool ok =0 ;
    while(sz(pq)){
      ll z = pq.top() ;pq.pop() ;
      if(ok)dp[v][0] += z ;
      ok = 1; 
      dp[v][1] += z; 
    }
  }
}

  vector<ll> ans ;

void sl(int l , int r , ll a =-1 , ll b= -1){
  if(a==-1){
    k = l ;dfs(0);
    a =dp[0][1];
  }
  if(b==-1){
    k = r ;dfs(0);
    b =dp[0][1];
  }
  if(a==b){
    rep(i , l , r)ans.pb(a);
  }else{
    int mid = (l+r)/2 ;
    sl(l , mid , a , -1);
    sl(mid+1,r,-1,b); 
  }
}

vector<ll> minimum_closure_costs(int n, vector<int> U,vector<int> V,vector<int> W) {
  rep(i ,0 , n-2){
    G[V[i]].pb({U[i] , W[i]}) ;
    G[U[i]].pb({V[i] , W[i]}) ;
  }
  sl(0 ,n-1);
  return ans ;
}

• Subtask 1: 0 / 5
• Subtask 2: 7 / 7
• Subtask 3: 14 / 14
• Subtask 4: 10 / 10
• Subtask 5: 0 / 17
• Subtask 6: 0 / 25
• Subtask 7: 0 / 22