제출 #1240682

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1240682		mychecksedad	Closing Time (IOI23_closing)	C++17	75 / 100	185 ms	12356 KiB

closing

#include "closing.h"
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define all(x) x.begin(),x.end()
#define vi vector<int>
#define pii pair<int,int>
#define ff first
#define ss second
#define ll long long int
#define cerr if(0) cerr
const int N = 5000, M = 3e5;

multiset<ll> S[N];
ll A[M], B[M], C[M]; // 0 - 1
vector<pair<int, ll>> g[M];
bitset<M> in_path;
int sz[M];
bool dfs(int v, int p, int t, vector<int> &T){
  T.pb(v);
  if(v == t){
    return 1;
  }
  for(auto [u, w]: g[v]){
    if(u != p){
      bool ok = dfs(u, v, t, T);
      if(ok) return true;
    }
  }
  T.pop_back();
  return false;
}

void dfs2(int v, int p){
  vi U;
  sz[v] = 1;
  int big = -1;
  for(auto [u, w]: g[v]){
    if(u != p && !in_path[u]){
      dfs2(u, v);
      sz[v] += sz[u];
      U.pb(u);
      if(big == -1 || sz[big] < sz[u]) big = u;
    }
  }
  if(big == -1){
    S[v].insert(A[v]);
    return;
  }
  S[v].swap(S[big]);
  for(int u: U){
    if(u != big){
      for(ll x: S[u]) S[v].insert(x);
    }
  }
  S[v].insert(A[v]);
}

int max_score(int n, int X, int Y, long long K, std::vector<int> U, std::vector<int> V, std::vector<int> W)
{
  for(int i = 0; i < n; ++i) g[i].clear(), S[i].clear();
  in_path = 0;
  for(int i = 0; i < n-1; ++i){
    g[U[i]].pb({V[i], W[i]});
    g[V[i]].pb({U[i], W[i]});
  }
  vector<ll> dist(n, -1);
  dist[X] = 0;
  queue<int> q;
  q.push(X);
  while(!q.empty()){
    int v = q.front(); q.pop();
    for(auto [u, w]: g[v]){
      if(dist[u] == -1){
        dist[u] = dist[v] + w;
        q.push(u);
      }
    }
  }
  vector<ll> dist2(n, -1);
  dist2[Y] = 0;
  q.push(Y);
  while(!q.empty()){
    int v = q.front(); q.pop();
    for(auto [u, w]: g[v]){
      if(dist2[u] == -1){
        dist2[u] = dist2[v] + w;
        q.push(u);
      }
    }
  }
  for(int i = 0; i < n; ++i){
    A[i] = min(dist[i], dist2[i]);
    B[i] = max(dist[i], dist2[i]) - min(dist[i], dist2[i]);
  }
  vi T; 
  dfs(X, X, Y, T);
  for(int x: T) in_path[x] = 1;

  ll sum = 0;
  for(int x: T){
    sum += A[x];
  }

  int ans = 0;  
  ll KK = K;
  for(int j = 0; j < n; ++j) C[j] = A[j];
  
  sort(C, C+n);
  for(int j = 0; j < n; ++j){
    if(KK>= C[j]) KK-=C[j], ++ans;
  }

  if(dist[Y] > 2*K) return ans;

  for(int x: T){
    K -= A[x];
    A[x] = 0;
  }

  vector<ll> DP(2*n + 5, 1ll*(1e18));

  DP[0] = 0;
  vector<ll> D;
  int cur = 0;
  for(int v: T){
    dfs2(v, v);
    vector<ll> dp(2*sz[v] + 5, 1ll*(1e18));
    dp[0] = 0;
    vector<ll> sum(sz[v] + 1);
    auto it = S[v].begin();
    for(int i = 1; i <= sz[v]; ++i, ++it){
      sum[i] = sum[i - 1] + (*it);
    }
    for(int i = 1; i <= 2*sz[v]; ++i){
      for(int j = 0; j * 2 <= i; ++j){
        if(i - j <= sz[v])
          dp[i] = min(j * B[v] + sum[i - j], dp[i]);
      }
    }


    for(int j = cur; j >= 0; --j){
      for(int w = 1; w <= 2*sz[v]; ++w){
        DP[j + w] = min(DP[j + w], DP[j] + dp[w]);
      }
    }
    cur += 2 * sz[v];
  }


  for(int i = 2*n; i >= 0; --i) if(DP[i] <= K) return max(i, ans);
  return ans;
}

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• Subtask 2: 9 / 9
• Subtask 3: 12 / 12
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