結果

問題 No.2634 Tree Distance 3
ユーザー umimel
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
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 * 69
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 #
プレゼンテーションモードにする

#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--){
//
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();
}
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