結果
問題 |
No.2634 Tree Distance 3
|
ユーザー |
|
提出日時 | 2024-02-16 23:01:34 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 881 ms / 3,000 ms |
コード長 | 4,136 bytes |
コンパイル時間 | 2,366 ms |
コンパイル使用メモリ | 193,148 KB |
実行使用メモリ | 55,116 KB |
最終ジャッジ日時 | 2024-09-28 21:36:59 |
合計ジャッジ時間 | 34,683 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 69 |
ソースコード
#include<bits/stdc++.h> using namespace std; using ll = long long; using pll = pair<ll, ll>; #define drep(i, cc, n) for (ll i = (cc); i <= (n); ++i) #define rep(i, n) drep(i, 0, n - 1) #define all(a) (a).begin(), (a).end() #define pb push_back #define fi first #define se second mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count()); const ll MOD1000000007 = 1000000007; const ll MOD998244353 = 998244353; const ll MOD[3] = {999727999, 1070777777, 1000000007}; const ll LINF = 1LL << 60LL; const int IINF = (1 << 30) - 1; template<typename T> struct edge{ int from, to; T cost; int id; edge(){} edge(int to, T cost=1) : from(-1), to(to), cost(cost){} edge(int to, T cost, int id) : from(-1), to(to), cost(cost), id(id){} edge(int from, int to, T cost, int id) : from(from), to(to), cost(cost), id(id){} }; template<typename T> struct redge{ int from, to; T cap, cost; int rev; redge(int to, T cap, T cost=(T)(1)) : from(-1), to(to), cap(cap), cost(cost){} redge(int to, T cap, T cost, int rev) : from(-1), to(to), cap(cap), cost(cost), rev(rev){} }; template<typename T> using Edges = vector<edge<T>>; template<typename T> using weighted_graph = vector<Edges<T>>; template<typename T> using tree = vector<Edges<T>>; using unweighted_graph = vector<vector<int>>; template<typename T> using residual_graph = vector<vector<redge<T>>>; template<typename W> struct lowest_common_ancestor{ int root = 0; int n; int log_n = 1; vector<vector<int>> par; vector<int> depth; lowest_common_ancestor(tree<W> &T, int root=0) : root(root){ n = (int)T.size(); while((1 << log_n) < n) log_n++; par.resize(log_n, vector<int>(n, -1)); depth.resize(n, 0); init(T); } void init(tree<W> &T){ function<void(int, int)> dfs = [&](int v, int p){ par[0][v] = p; for(edge<W> e : T[v]) if(e.to!=p){ depth[e.to] = depth[v]+1; dfs(e.to, v); } }; dfs(root, -1); for(int k=0; k+1<log_n; k++){ for(int v=0; v<n; v++){ if(par[k][v]<0) par[k+1][v] = -1; else par[k+1][v] = par[k][par[k][v]]; } } } int query(int u, int v){ if(depth[u] < depth[v]) swap(u, v); for(int k=0; k<log_n; k++) if((depth[u]-depth[v]) >> k & 1) u = par[k][u]; if(u == v) return u; for(int k=log_n-1; k>=0; k--){ if(par[k][u] != par[k][v]){ u = par[k][u]; v = par[k][v]; } } return par[0][u]; } }; void solve(){ int n; cin >> n; vector<int> a(n); for(int i=0; i<n; i++) cin >> a[i]; { vector<int> c(n); for(int i=0; i<n; i++) c[i] = a[i]; sort(all(c)); c.erase(unique(all(c)), c.end()); for(int i=0; i<n; i++) a[i] = lower_bound(all(c), a[i])-c.begin(); } tree<int> T(n); for(int i=0; i<n-1; i++){ int u, v; cin >> u >> v; u--; v--; T[u].pb(edge<int>(v)); T[v].pb(edge<int>(u)); } lowest_common_ancestor<int> lca(T); vector<int> depth(n, 0); function<void(int, int)> dfs = [&](int v, int p){ for(edge<int> e : T[v]) if(e.to!=p){ depth[e.to] = depth[v]+1; dfs(e.to, v); } }; dfs(0, -1); function<int(int, int)> dist = [&](int u, int v){ return depth[u] + depth[v] -2*depth[lca.query(u, v)]; }; vector<vector<int>> b(n); for(int i=0; i<n; i++) b[a[i]].push_back(i); int s = -1, t = -1; int diam = 0; vector<int> ans(n, 0); for(int i=n-1; i>=0; i--){ //追加 for(int v : b[i]){ if(s==-1){ s = v; t = v; diam = 0; }else{ int s_dist = dist(s, v); int t_dist = dist(t, v); if(diam < max(s_dist, t_dist)){ if(s_dist>=t_dist){ t = v; diam = s_dist; }else{ s = v; diam = t_dist; } } } } for(int v : b[i]){ ans[v] = max(dist(s, v), dist(t, v)); } } for(int i=0; i<n; i++) cout << ans[i] << " "; cout << '\n'; } int main(){ cin.tie(nullptr); ios::sync_with_stdio(false); int T=1; //cin >> T; while(T--) solve(); }