Submission #433559

#	Time	Username	Problem	Language	Result	Execution time	Memory
433559		misir	Palindromes (APIO14_palindrome)	C++14	100 / 100	99 ms	73800 KiB

palindrome

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
using pii = pair<int, int>;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;
//const ll mod = 998244353;


inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }



int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int dx8[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dy8[] = {1, -1, -1, 0, 1, -1, 0, 1};
int kx8[] = {1, 1, 2, 2, -1, -1, -2, -2};
int ky8[] = {2, -2, 1, -1, 2, -2, 1, -1};
/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
//*/
///**
template < typename F, typename S >ostream& operator << ( ostream& os, const pair< F, S > & p ) {return os << "(" << p.first << ", " << p.second << ")";}
template < typename T >ostream &operator << ( ostream & os, const vector< T > &v ) {os << "{"; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << *it;} return os << "}";}
template < typename T >ostream &operator << ( ostream & os, const set< T > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin()) os << ", "; os << *it;} return os << "]";}
template < typename F, typename S >ostream &operator << ( ostream & os, const map< F, S > &v ) {os << "["; for (auto it = v.begin(); it != v.end(); ++it) {if ( it != v.begin() ) os << ", "; os << it -> first << " = " << it -> second ;} return os << "]";}
#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)
clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)
void faltu () { cerr << endl; }
template <typename T>void faltu( T a[], int n ) {for (int i = 0; i < n; ++i) cerr << a[i] << ' '; cerr << endl;}
template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }

// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
using namespace __gnu_pbds;

// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;
// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।
//*//**___________________________________________________**/
const int N = 303030;

struct Etree
{
    int nxt[26], sufflink;
    ll len, cnt;
    vector<int>edges;
} Tree[N];
string s;
int suff, num;
ll ans = 0;
void add_letter(int pos) {
    int cur = suff, cur_len = 0;
    int letter = s[pos] - 'a';

    while (true) {
        cur_len = Tree[cur].len;
        if (pos - 1 - cur_len > -1 && s[pos - 1 - cur_len] == s[pos])
            break;
        cur = Tree[cur].sufflink;
    }
    if (Tree[cur].nxt[letter]) {
        suff = Tree[cur].nxt[letter];
        Tree[suff].cnt++;
        return;
    }
    suff = ++num;
    Tree[num].len = Tree[cur].len + 2;
    Tree[num].cnt = 1;
    Tree[cur].nxt[letter] = num;

    if (Tree[num].len == 1) {
        Tree[num].sufflink = 2;
        Tree[2].edges.push_back(num);
        return;
    }

    while (true) {
        cur = Tree[cur].sufflink;
        cur_len = Tree[cur].len;

        if (pos - 1 - cur_len > -1 && s[pos - 1 - cur_len] == s[pos]) {
            Tree[num].sufflink = Tree[cur].nxt[letter];
            Tree[Tree[cur].nxt[letter]].edges.push_back(num);
            break;
        }
    }
}

void init() {
    num = suff = 2;
    Tree[1].len = -1, Tree[1].sufflink = 1;
    Tree[2].len = 0, Tree[2].sufflink = 1;
    Tree[1].edges.push_back(2);
}

void dfs(int u = 1) {
    for (int v : Tree[u].edges) {
        dfs(v);
        Tree[u].cnt += Tree[v].cnt;
    }
    ans = max(ans, Tree[u].len * Tree[u].cnt);
}

int main()
{
    FASTIO
    /*
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    freopen("error.txt", "w", stderr);
#endif
//*/
    cin >> s;
    // dbg(s);
    init();
    int n = s.size();
    // dbg(n);
    for (int i = 0; i < n; i++)add_letter(i);
    dfs();
    cout << ans << "\n";
    return 0;
}

• Subtask 1: 8 / 8
• Subtask 2: 15 / 15
• Subtask 3: 24 / 24
• Subtask 4: 26 / 26
• Subtask 5: 27 / 27