Submission #1135032

#	Time	Username	Problem	Language	Result	Execution time	Memory
1135032		pan	Palindromi (COCI22_palindromi)	C++20	110 / 110	383 ms	195760 KiB

Main

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

struct errtree
{
	ll ans, nxtl, nxtr, _size;
	vector<ll> len, dp[2], to[2];
	deque<ll> arr;
	errtree()
	{
		len.pb(-1); len.pb(0); // dummy
		nxtl = nxtr = 1; _size = 0, ans = 2;
		for (ll j=0; j<2; ++j) for (ll i=0; i<2; ++i) {dp[i].pb(0); to[i].pb(0);}
		
	}
	void insert(ll from, ll val, ll back)
	{
		len.pb(len[from] + 2);
		for (ll i=0; i<2; ++i) to[i].pb(0);
		to[val][from] = ans++;
		ll prevv = ((from==0)?1:to[val][dp[val][from]]);
		ll val2 = 0;
		if (back) val2 = arr[_size - len[prevv]];
		else val2 = arr[len[prevv]];
		dp[val2].pb(prevv);
		dp[val2^1].pb(dp[val2^1][prevv]);
	}
	void insert_right(ll val)
	{
		arr.pb(val);
		ll now;
		if (len[nxtr] < _size && arr[_size - len[nxtr] - 1] == val) now = nxtr;
		else now = dp[val][nxtr];
		if (to[val][now]==0) insert(now, val, 1);
		++_size;
		nxtr = to[val][now];
		if (len[nxtr]==_size) nxtl = nxtr;
	}
	void insert_left(ll val)
	{
		ll now;
		if (len[nxtl] < _size &&arr[len[nxtl]] == val) now = nxtl;
		else now = dp[val][nxtl];
		arr.push_front(val);
		if (to[val][now]==0) insert(now, val, 0);
		++_size;
		nxtl= to[val][now];
		if (len[nxtl]==_size) nxtr = nxtl;
	}
}root[100005];

// UFDS
ll par[100005];
ll find(ll x) {return ((x==par[x])?x:par[x] = find(par[x]));}

int main()
{
	ll n;
	string ss;
	cin >> n;
	cin >> ss;
	for (ll i=0; i<n; ++i) par[i] = i;
	for (ll i=0; i<n; ++i) root[i].insert_left(ss[i]-'0');
	for (ll i=0; i<n-1; ++i) 
	{
		ll u, v;
		cin >> u >> v; u--; v--;
		u = find(u); v = find(v);
		if (root[u]._size < root[v]._size)
		{
			for (ll j=0; j<root[u]._size; ++j)
			{
				root[v].insert_left(root[u].arr[root[u]._size-j-1]);
			}
			cout << root[v].ans-2 << endl;
			par[u] = v;
		}
		else
		{
			for (ll j=0; j<root[v]._size; ++j)
			{
				root[u].insert_right(root[v].arr[j]);
			}
			cout << root[u].ans-2 << endl;
			par[v] = u;
		}
	}
	return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 30 / 30
• Subtask 4: 50 / 50