結果

問題 No.2634 Tree Distance 3
コンテスト
ユーザー ponjuice
提出日時 2024-02-16 23:07:33
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 7,296 bytes
コンパイル時間 4,465 ms
コンパイル使用メモリ 264,552 KB
最終ジャッジ日時 2025-02-19 14:41:12
ジャッジサーバーID
（参考情報） 		judge1 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 57 WA * 12
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
using namespace atcoder;

//高速化 
struct ponjuice{ponjuice(){cin.tie(0);ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);}}PonJuice;
//#define endl '\n' //インタラクティブ問題の時は消す

//型
using ll = long long;
using ld = long double;
using mint = modint998244353;//1000000007;
template<class T>using vc = vector<T>; template<class T>using vvc = vc<vc<T>>; template<class T>using vvvc = vvc<vc<T>>;
using vi = vc<int>;  using vvi = vvc<int>;  using vvvi = vvvc<int>;
using vl = vc<ll>;   using vvl = vvc<ll>;   using vvvl = vvvc<ll>;
using pi = pair<int, int>;  using pl = pair<ll, ll>;
using ull = unsigned ll;
template<class T>using priq = priority_queue<T>;
template<class T>using priqg = priority_queue<T, vc<T>, greater<T>>;

// for文
#define overload4(a, b, c, d, e, ...) e
#define rep1(n)             for(ll i = 0; i < n; i++)
#define rep2(i, n)          for(ll i = 0; i < n; i++)
#define rep3(i, a, b)       for(ll i = a; i < b; i++)
#define rep4(i, a, b, step) for(ll i = a; i < b; i+= step)
#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)
#define per1(n)             for(ll i = n-1; i >= 0; i--)
#define per2(i, n)          for(ll i = n-1; i >= 0; i--)
#define per3(i, a, b)       for(ll i = b-1; i >= a; i--)
#define per4(i, a, b, step) for(ll i = b-1; i >= a; i-= step)
#define per(...) overload4(__VA_ARGS__, per4, per3, per2, per1)(__VA_ARGS__)
#define fore1(a)       for(auto&& i : a)	
#define fore2(i,a)     for(auto&& i : a)
#define fore3(x,y,a)   for(auto&& [x, y] : a)
#define fore4(x,y,z,a) for(auto&& [x, y, z] : a)
#define fore(...) overload4(__VA_ARGS__, fore4, fore3, fore2, fore1)(__VA_ARGS__)

//関数
#define mp make_pair
#define mt make_tuple
#define a first
#define b second
#define pb emplace_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define si(x) (ll)(x).size()
template<class S, class T>inline bool chmax(S& a, T b){return a < b && ( a = b , true);}
template<class S, class T>inline bool chmin(S& a, T b){return a > b && ( a = b , true);}
template<class T>void uniq(vc<T>&a){sort(all(a));a.erase(unique(all(a)),a.end());}
template<class T>vc<T> operator++(vc<T>&v,signed){auto res = v;fore(e,v)e++;return res;}
template<class T>vc<T> operator--(vc<T>&v,signed){auto res = v;fore(e,v)e--;return res;}
template<class T>vc<T> operator++(vc<T>&v){fore(e,v)e++;return v;}
template<class T>vc<T> operator--(vc<T>&v){fore(e,v)e--;return v;}

//入出力(operator)
template<int T>istream&operator>>(istream&is,static_modint<T>&a){ll v;is>>v;a=v;return is;}
istream&operator>>(istream&is,modint&a){ll v;cin>>v;a=v;return is;}
template<class S,class T>istream&operator>>(istream&is,pair<S,T>&a){is>>a.a>>a.b;return is;}
template<class T>istream&operator>>(istream&is,vc<T>&a){fore(e,a)is>>e;return is;}

template<int T>ostream&operator<<(ostream&os,static_modint<T>a){return os<<a.val();}
ostream&operator<<(ostream&os,modint a){return os<<a.val();}
template<class S,class T>ostream&operator<<(ostream&os,pair<S,T>&a){return os<<a.a<<" "<<a.b;}
template<class T>ostream&operator<<(ostream&os,set<T>&a){fore(it,a){os<<it<<" ";}return os;}
template<class T>ostream&operator<<(ostream&os,multiset<T>&a){fore(it,a){os<<it<<" ";}return os;}
template<class S,class T>ostream&operator<<(ostream&os,map<S,T>&a){fore(x,y,a){os<<x<<" "<<y<<"\n";}return os;}
template<class T>ostream&operator<<(ostream&os,unordered_set<T>&a){fore(it,a){os<<it<<" ";}return os;}
template<class S,class T>ostream&operator<<(ostream&os,unordered_map<S,T>&a){fore(x,y,a){os<<x<<" "<<y<<"\n";}return os;}
template<class T>ostream&operator<<(ostream&os,vc<T>&a){fore(e,a)os<<e<<" ";return os;}
template<class T>ostream&operator<<(ostream&os,vvc<T>&a){fore(e,a)os<<e<<"\n";return os;}

//入出力(関数)
vi readvi(ll n){vi a(n);cin>>a;return a;}
vl readvl(ll n){vl a(n);cin>>a;return a;}
vvi readg(ll n,ll m,bool bidirected=true){vvi g(n);rep(i,m){ll a,b;cin>>a>>b;a--;b--;g[a].pb(b);if(bidirected)g[b].pb(a);}return g;}
vvc<pi>readgc(ll n,ll m,bool bidirected=true){vvc<pi> g(n);rep(i,m){ll a,b,c;cin>>a>>b>>c;a--;b--;g[a].pb(b,c);if(bidirected)g[b].pb(a,c);}return g;}
vvi readt(ll n,bool bidirected=true){return readg(n,n-1,bidirected);}
vvc<pi> readtc(ll n,bool bidirected=true){return readgc(n,n-1,bidirected);}

inline void yes(){cout << "Yes\n";}
inline void no(){cout << "No\n";}
inline void yesno(bool y = true){if(y)yes();else no();}
inline void print(){cout<<endl;}
template<class T>inline void print(T a){cout<<a<<endl;}
template<class T,class... Ts>inline void print(T a,Ts ...b){cout<<a;(cout<<...<<(cout<<' ',b));cout<<endl;}

//定数
constexpr ll mod = 998244353;
constexpr ll minf=-(1<<29);
constexpr ll inf=(1<<29);
constexpr ll MINF=-(1LL<<60);
constexpr ll INF=(1LL<<60);
constexpr ld EPS = 1e-8;
const ld PI = acosl(-1);
#define equals(a, b) (abs((a) - (b)) < EPS)
const int dx[4] ={-1, 0, 1, 0};
const int dy[4] ={ 0, 1, 0,-1};
const int dx8[8] ={-1,-1,-1, 0, 1, 1, 1, 0};
const int dy8[8] ={-1, 0, 1, 1, 1, 0,-1,-1};

void solve();
int main() {
	int t = 1;
    // cin >>t;
    while(t--)solve();
}

class LCA{
  private:
  int root;
  int k; // n<=2^kとなる最小のk
  vector<vector<int>> dp; // dp[i][j]:=要素jの2^i上の要素
  vector<int> depth;  // depth[i]:=rootに対する頂点iの深さ
  
  public:
  LCA(const vector<vector<int>>& _G, const int _root=0){
    int n=_G.size();
    root=_root;
    k=1;
    int nibeki=2;
    while(nibeki<n){
      nibeki<<=1;
      k++;
    }
    // 頂点iの親ノードを初期化
    dp = vector<vector<int>>(k+1, vector<int>(n, -1));
    depth.resize(n);
    function<void(int, int)> _dfs=[&](int v, int p){
      dp[0][v]=p;
      for(auto nv: _G[v]){
        if(nv==p) continue;
        depth[nv]=depth[v]+1;
        _dfs(nv, v); 
      }
    };
    _dfs(root, -1);
    // ダブリング
    for(int i=0; i<k; i++){
      for(int j=0; j<n; j++){
        if(dp[i][j]==-1) continue;
        dp[i+1][j]=dp[i][dp[i][j]];
      }
    }
  }
  
  /// get LCA
  int get(int u, int v){
    if(depth[u]<depth[v]) swap(u,v); // u側を深くする
    if(depth[u]!=depth[v]){
      long long d=depth[u]-depth[v];
      for(int i=0; i<k; i++) if((d>>i)&1) u=dp[i][u];
    }
    if(u==v) return u;
    
    for(int i=k; i>=0; i--){
      if(dp[i][u]!=dp[i][v]){
        u=dp[i][u], v=dp[i][v];
      }
    }
    return dp[0][u];
  }
  
  int get_distance(const int u, const int v){
    int lca=get(u,v);
    return depth[u]+depth[v]-2*depth[lca];
  }
};

void solve(){
    int n;
    cin >> n;
    vi a = readvi(n);
    vvi t = readt(n);
    LCA dist(t);
    vi io(n);
    iota(all(io), 0);
    sort(all(io), [&](int l,int r){
        return a[l] > a[r];
    });
    vi st;
    vi ans(n);
    int l = io[0],r = io[0];
    rep(k,n){
        int i = io[k];
        if(k == 0 || a[i] != a[io[k-1]]){
            rep(L, k, n){
                int j = io[L];
                if(a[i] != a[j]) break; 
                if(dist.get_distance(j,r) > dist.get_distance(l,r)) l = j;
                if(dist.get_distance(j,l) > dist.get_distance(l,r)) r = j;
            }
        }
        ans[i] = max(dist.get_distance(i,l),dist.get_distance(i,r));
    }

    print(ans);
}
0