Submission #954362

#	Time	Username	Problem	Language	Result	Execution time	Memory
954362		GrindMachine	Meetings 2 (JOI21_meetings2)	C++17	100 / 100	488 ms	50880 KiB

meetings2

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
 
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
 
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)
 
template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}
 
template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}
 
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
 
/*
 
refs:
edi

odd c => ans = 1
assume c is even
let's say all nodes on the path (u,v) are good
then root the tree on the path (u,v)
path is good if the subtrees of u and v both have size >= c/2
(cuz only then c/2 nodes can be put into the subtrees of u and v)
 
however, if u is the ances of v (or vice versa), the condition checks become a little tedious
same condition still holds, but can't just check subtree sizes (cuz one is contained in another)

key idea:
root the tree at the centroid

this is a clean fix, because now (u,v) s.t u is an ances of v is unoptimal
(u,v) for u != centroid is suboptimal, because we can always do (centroid,v) (recall centroid properties)
no more corner cases to worry about

activate nodes in dec ord of subtree size
every time node is activated, find the farthest distance to another activated node
can be done by maintaining the diameter of the active tree (active tree always forms a connected subgraph) and updating as new nodes are activated

*/
 
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
 
vector<ll> adj[N];
 
struct lca_algo {
    // LCA template (for graphs with 1-based indexing)
 
    int LOG = 1;
    vector<int> depth;
    vector<vector<int>> up;
    vector<int> tin, tout;
    int timer = 1;
 
    lca_algo() {
 
    }
 
    lca_algo(int n) {
        lca_init(n);
    }
 
    void lca_init(int n) {
        while ((1 << LOG) < n) LOG++;
        up = vector<vector<int>>(n + 1, vector<int>(LOG, 1));
        depth = vector<int>(n + 1);
        tin = vector<int>(n + 1);
        tout = vector<int>(n + 1);
 
        lca_dfs(1, -1);
    }
 
    void lca_dfs(int node, int par) {
        tin[node] = timer++;
 
        trav(child, adj[node]) {
            if (child == par) conts;
 
            up[child][0] = node;
            rep1(j, LOG - 1) {
                up[child][j] = up[up[child][j - 1]][j - 1];
            }
 
            depth[child] = depth[node] + 1;
 
            lca_dfs(child, node);
        }
 
        tout[node] = timer-1;
    }
 
    int lift(int u, int k) {
        rep(j, LOG) {
            if (k & (1 << j)) {
                u = up[u][j];
            }
        }
 
        return u;
    }
 
    int query(int u, int v) {
        if (depth[u] < depth[v]) swap(u, v);
        int k = depth[u] - depth[v];
        u = lift(u, k);
 
        if (u == v) return u;
 
        rev(j, LOG - 1, 0) {
            if (up[u][j] != up[v][j]) {
                u = up[u][j];
                v = up[v][j];
            }
        }
 
        u = up[u][0];
        return u;
    }
 
    int get_dis(int u, int v) {
        int lca = query(u, v);
        return depth[u] + depth[v] - 2 * depth[lca];
    }
 
    bool is_ances(int u, int v){
        return tin[u] <= tin[v] and tout[u] >= tout[v];
    }
};
 
vector<ll> subsiz(N), depth(N);
 
void dfs1(ll u, ll p){
    subsiz[u] = 1;
    trav(v,adj[u]){
        if(v == p) conts;
        depth[v] = depth[u]+1;
        dfs1(v,u);
        subsiz[u] += subsiz[v];
    }
}
 
ll dfs2(ll u, ll p){
    trav(v,adj[u]){
        if(v == p) conts;
        if(subsiz[v] > subsiz[1]/2){
            return dfs2(v,u);
        }
    }
  
    return u;
}
 
void solve(int test_case)
{
    ll n; cin >> n;
    rep1(i,n-1){
        ll u,v; cin >> u >> v;
        adj[u].pb(v), adj[v].pb(u);
    }
 
    dfs1(1,-1);
    ll r = dfs2(1,-1);    
    dfs1(r,-1);
 
    lca_algo LCA(n);
 
    vector<pll> order;
    rep1(i,n) order.pb({subsiz[i],i});
    sort(rall(order));
    vector<ll> ans(n+5);
    array<ll,3> diam = {0,r,r};
 
    for(auto [siz,u] : order){
        auto nxt_diam = diam;
        rep1(j,2){
            ll v = diam[j];
            array<ll,3> ar = {LCA.get_dis(u,v),u,v};
            amax(nxt_diam,ar);
        }
 
        diam = nxt_diam;
 
        rep1(j,2){
            ll v = diam[j];
            ll d = LCA.get_dis(u,v);
            amax(ans[siz],d);
        }
    }
 
    rev(i,n,1) amax(ans[i],ans[i+1]);
 
    rep1(i,n){
        if(i&1) cout << 1 << endl;
        else cout << ans[i/2]+1 << endl;
    }
}
 
int main()
{
    fastio;
 
    int t = 1;
    // cin >> t;
 
    rep1(i, t) {
        solve(i);
    }
 
    return 0;
}

• Subtask 1: 4 / 4
• Subtask 2: 16 / 16
• Subtask 3: 80 / 80