Submission #1150913

#	Time	Username	Problem	Language	Result	Execution time	Memory
1150913		jiahng	Cat Exercise (JOI23_ho_t4)	C++20	54 / 100	136 ms	74056 KiB

Main

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

typedef long long ll;
#define ll int
typedef pair<int,int> pi;
typedef pair<ll,ll> pill;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi, ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define mp make_pair
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define INF (ll)1e18
#define MOD 1000000007
typedef pair <vi, int> pvi;
typedef pair <int,pi> ipi;
typedef vector <pii> vpii;
typedef pair <pi,pi> pipi;
#define maxn 500010

int N;
int dp[maxn],A[maxn],B[maxn],best[maxn];
vi adj[maxn];
int p[maxn],twok[maxn][20],depth[maxn];
int fl(int x){
	if (p[x] == x) return x;
	return p[x] = fl(p[x]);
}
void dfs(int x,int p){
	twok[x][0] = p;
	FOR(i,1,19){
		if (twok[x][i-1] == -1) break;
		twok[x][i] = twok[twok[x][i-1]][i-1];
	}
	aFOR(i, adj[x]) if (i != p){
		depth[i] = depth[x] + 1;
		dfs(i,x);
	}
}
int lca(int a,int b){
	if (depth[a] < depth[b]) swap(a,b);
	int k = depth[a] - depth[b];
	FOR(i,0,19) if (k & (1 << i)){
		a = twok[a][i];
	}
	
	if (a == b) return a;
	DEC(i,19,0) if (twok[a][i] != twok[b][i]){
		a = twok[a][i]; b = twok[b][i];
	}
	return twok[a][0];
}

int d(int a,int b){
	int c = lca(a,b);
	return depth[a] + depth[b] - 2 * depth[c];
}
int32_t main(){
	fast;
	cin >> N;
	FOR(i,1,N) cin >> A[i];
	FOR(i,1,N) B[A[i]] = i;
	int a,b;
	FOR(i,1,N-1){
		cin >> a >> b;
		adj[a].pb(b); adj[b].pb(a);
	}
	mem(twok,-1);
	dfs(1,-1);
	FOR(i,1,N) p[i] = i;
	FOR(i,1,N){
		int x = B[i];
		aFOR(j, adj[x]) if (A[j] < i){
			int opt = fl(j);
			dp[x] = max(dp[x],d(opt,x) + dp[opt]); 
			p[opt] = x;
		}		
	}
	
	cout << dp[B[N]];
}

• Subtask 1: 7 / 7
• Subtask 2: 7 / 7
• Subtask 3: 7 / 7
• Subtask 4: 10 / 10
• Subtask 5: 0 / 20
• Subtask 6: 23 / 23
• Subtask 7: 0 / 26