Submission #392336

#	Time	Username	Problem	Language	Result	Execution time	Memory
392336		BorisBarca	Palinilap (COI16_palinilap)	C++14	37 / 100	112 ms	28744 KiB

palinilap

/*
$$$$$$$\                      $$\           $$$$$$$\
$$  __$$\                     \__|          $$  __$$\
$$ |  $$ | $$$$$$\   $$$$$$\  $$\  $$$$$$$\ $$ |  $$ | $$$$$$\   $$$$$$\   $$$$$$$\ $$$$$$\
$$$$$$$\ |$$  __$$\ $$  __$$\ $$ |$$  _____|$$$$$$$\ | \____$$\ $$  __$$\ $$  _____|\____$$\
$$  __$$\ $$ /  $$ |$$ |  \__|$$ |\$$$$$$\  $$  __$$\  $$$$$$$ |$$ |  \__|$$ /      $$$$$$$ |
$$ |  $$ |$$ |  $$ |$$ |      $$ | \____$$\ $$ |  $$ |$$  __$$ |$$ |      $$ |     $$  __$$ |
$$$$$$$  |\$$$$$$  |$$ |      $$ |$$$$$$$  |$$$$$$$  |\$$$$$$$ |$$ |      \$$$$$$$\\$$$$$$$ |
\_______/  \______/ \__|      \__|\_______/ \_______/  \_______|\__|       \_______|\_______|
*/
#include <bits/stdc++.h>
#include <ext/pb_ds/detail/standard_policies.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

#define PB push_back
#define MP make_pair
#define INS insert
#define LB lower_bound
#define UB upper_bound
#define pii pair <int,int>
#define pll pair <long long, long long>
#define X first
#define Y second
#define _ << " " <<
#define sz(x) (int)x.size()
#define all(a) (a).begin(),(a).end()
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORD(i, a, b) for (int i = (a); i > (b); --i)
#define FORA(i, x) for (auto &i : x)
#define REP(i, n) FOR(i, 0, n)
#define BITS(x) __builtin_popcount(x)
#define SQ(a) (a) * (a)
#define TRACE(x) cout << #x " = " << (x) << '\n';
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define umap unordered_map

typedef long long ll;
typedef long double ld;
typedef vector <int> vi;
typedef vector <pii> vpi;
typedef vector <ll> vll;
typedef vector <pll> vpl;
typedef vector <double> vd;
typedef vector <ld> vld;
typedef vector<string> vs;
typedef tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
//find_by_order -> kti najmanji
//order_of_key -> koliko ima manjih od x
//((float) t)/CLOCKS_PER_SEC

const int MOD = 1e9 + 7;
const double PI = acos(-1);
const int INF = 1e9 + 10;
const ll INFL = 1e18 + 10;
const int ABC = 30;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dox[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int doy[] = {0, 0, -1, 1, -1, 1, -1, 1};

inline int sum(int a, int b){
	if (a + b < 0)
		return (a + b + MOD) % MOD;
	return (a + b) % MOD;
}

inline void add(int &a, int b){
	a = sum(a, b);
}

inline int mul(ll a, ll b){
	return ((a % MOD) * ((ll)b % MOD)) % MOD;
}

inline int sub(int a, int b){
	return (a - b + MOD) % MOD;
}

inline int fast_pot(ll pot, ll n){
	ll ret = 1;
	while (n){
		if (n & 1LL)
			ret = (ret * pot) % MOD;
		pot = (pot * pot) % MOD;
		n >>= 1LL;
	}
	return ret;
}

inline int divide(int a, int b){
	return mul(a, fast_pot(b, MOD - 2));
}

ll lcm(ll a, ll b){
	return abs(a * b) / __gcd(a, b);
}

inline int ccw(pii a, pii b, pii c) {
   return a.X * (b.Y - c.Y) + b.X * (c.Y - a.Y) + c.X * (a.Y - b.Y);
}


inline int CCW(pii A, pii B, pii C){
	double val = ccw(A, B, C);
	double eps = max(max(abs(A.X), abs(A.Y)), max(max(abs(B.X), abs(B.Y)), max(abs(C.X), abs(C.Y)))) / 1e9;
	if (val <= -eps)
		return -1;
	if (val >= eps)
		return 1;
	return 0;
}

void to_upper(string &x){
	REP(i, sz(x))
		x[i] = toupper(x[i]);
}

void to_lower(string &x){
	REP(i, sz(x))
		x[i] = tolower(x[i]);
}

string its(ll x){
	if (x == 0)
		return "0";
	string ret = "";
	while (x > 0){
		ret += (x % 10) + '0';
		x /= 10;
	}
	reverse(all(ret));
	return ret;
}

ll sti(string s){
	ll ret = 0;
	REP(i, sz(s)){
		ret *= 10;
		ret += (s[i] - '0');
	}
	return ret;
}

const int N = 1e5 + 10;
const int P = 31;

struct hsh{
	int h[N], p[N], n;
	
	void init(string s){
		h[0] = s[0] - 'a' + 1;
		p[0] = 1;
		n = sz(s);
		FOR(i, 1, n){
			h[i] = sum(mul(h[i - 1], P), s[i] - 'a' + 1);
			p[i] = mul(p[i - 1], P);
		}
	}

	int get(int i, int l){
		return sub(h[i + l - 1], (l ? mul(h[i - 1], p[l]) : 0));
	}

} h, rh;

int n;
ll dp[N][30], l[N], r[N], cntL[N], cntR[N];
string s;

ll solve(){
	int ret = 0;
	ll sol = n;
	FOR(i, 1, n - 1){
		int lo = 0, hi;
		if (s[i - 1] == s[i + 1]){
			lo = 1, hi = min(i, n - i - 1);
			while (lo < hi){
				int mid = (lo + hi + 1) / 2;
				//cout << mid _ s.substr(i - mid, mid) _ s.substr(i + 1, mid) _ h.get(i - mid, mid) _ rh.get(n - (i + mid) - 1, mid)  << '\n';
				if (h.get(i - mid, mid) == rh.get(n - (i + mid) - 1, mid))
					lo = mid;
				else
					hi = mid - 1;
			}
			sol += lo;
			l[i - 1] += lo;
			if (i - lo - 1 >= 0)
				cntL[i - lo - 1]--;
			cntL[i - 1]++;
			r[i + 1] += lo;
			if (i + lo + 1 < n)
				cntR[i + lo + 1]--;
			cntR[i + 1]++;

		}
		lo++;
		if (i - lo < 0 || i + lo >= n)
			continue;
		if (i - lo - 1 < 0 || i + lo + 1 >= n){
			dp[i - lo][s[i + lo] - 'a']++;
			dp[i + lo][s[i - lo] + 'a']++;
			continue;
		}
		int l = i - lo - 1;
		int r = i + lo + 1;
		//cout << l _ r << '\n';
		int lo1 = 0, hi1 = min(l, n - r - 1) + 1;
		while (lo1 < hi1){
			int mid = (lo1 + hi1 + 1) / 2;
			if (h.get(l - mid + 1, mid) == rh.get(n - (r + mid - 1) - 1, mid))
				lo1 = mid;
			else
				hi1 = mid - 1;
		}
		dp[l + 1][s[r - 1] - 'a'] += lo1 + 1;
		dp[r - 1][s[l + 1] - 'a'] += lo1 + 1;
		//cout << i _ lo1 << '\n';
	}

	FOR(i, 0, n - 1){
		int lo = 0, hi;
		if (s[i] == s[i + 1]){
			lo = 1, hi = min(i + 1, n - i - 1);
			while (lo < hi){
				int mid = (lo + hi + 1) / 2;
				if (h.get(i - mid + 1, mid) == rh.get(n - (i + mid) - 1, mid))
					lo = mid;
				else
					hi = mid - 1;
			}
			sol += lo;
			l[i] += lo;
			cntL[i]++;
			if (i - lo >= 0)
				cntL[i - lo]--;
			r[i + 1] += lo;
			cntR[i + 1]++;
			if (i + lo + 1 < n)
				cntR[i + lo + 1]--;
		}
		lo++;
		if (i - lo + 1 < 0 || i + lo >= n)
			continue;
		if (i - lo < 0 || i + lo >= n){
			dp[i - lo + 1][s[i + lo] - 'a']++;
			dp[i + lo][s[i - lo + 1] + 'a']++;
			continue;
		}
		int l = i - lo;
		int r = i + lo + 1;
		//cout << l _ r << '\n';
		int lo1 = 0, hi1 = min(l, n - r - 1) + 1;
		while (lo1 < hi1){
			int mid = (lo1 + hi1 + 1) / 2;
			if (h.get(l - mid + 1, mid) == rh.get(n - (r + mid - 1) - 1, mid))
				lo1 = mid;
			else
				hi1 = mid - 1;
		}
		//cout << i _ lo1 << '\n';
		dp[l + 1][s[r - 1] - 'a'] += lo1 + 1;
		dp[r - 1][s[l + 1] - 'a'] += lo1 + 1;
	}

	int tmp = 0, dec = 0;
	FORD(i, n - 1, -1){
		tmp += l[i];
		dec += cntL[i];
		REP(j, 26)
			if (j != s[i] - 'a')
				dp[i][j] -= tmp;
		tmp -= dec;
	}

	tmp = 0, dec = 0;
	REP(i, n){
		tmp += r[i];
		dec += cntR[i];
		//dp[i] -= tmp;
		REP(j, 26)
			if (j != s[i] - 'a')
				dp[i][j] -= tmp;
		tmp -= dec;
	}

	ll poc = sol;
	REP(i, n)
		REP(j, 26)
			sol = max(sol, poc + dp[i][j]);

			
	return sol;	
}


int main () {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	cin >> s;	
	n = sz(s);
	h.init(s);
	reverse(all(s));
	rh.init(s);
	reverse(all(s));
	cout << solve() << '\n';

	return 0;
}

Compilation message (stderr)

palinilap.cpp: In function 'll solve()':
palinilap.cpp:176:6: warning: unused variable 'ret' [-Wunused-variable]
  176 |  int ret = 0;
      |      ^~~

• Subtask 1: 0 / 17
• Subtask 2: 37 / 37
• Subtask 3: 0 / 46