제출 #1019911

#제출 시각아이디문제언어결과실행 시간메모리
1019911andrei_iorgulescu회문 (APIO14_palindrome)C++14
0 / 100
1014 ms21396 KiB
#include <bits/stdc++.h> using namespace std; #define int long long int n; char a[300005]; int modulo = 1e9 + 321; int base = 59; int powbase[300005],invpowbase[300005]; int h[300005],revh[300005];///sum din a[i] * B^(i - 1) int lgpow(int x,int y) { int z = 1; while (y != 0) { if (y % 2 == 1) z = z * x % modulo; x = x * x % modulo; y /= 2; } return z; } vector<double> ev[300005];///cu + daca il bag, - daca il scot bool pal(int l,int r) { if (l < 1 or r > n or l > r) return false; for (int i = l; i <= r; i++) if (a[i] != a[r + l - i]) return false; return true; int h1 = (h[r] - h[l - 1] + modulo) % modulo * invpowbase[l - 1] % modulo; int h2 = (revh[l] - revh[r + 1] + modulo) % modulo * invpowbase[n - r] % modulo; if (h1 == h2) return true; return false; } int get_hash(int l, int r) { int h1 = (h[r] - h[l - 1] + modulo) % modulo * invpowbase[l - 1] % modulo; return h1; } int lft[300005]; map<int,int> id; pair<int,int> what[300005];///sa stiu si eu cam cum arata un string de acel id int cnt_id; int f[300005]; int l2[300005]; signed main() { string when_what_where_how_why; cin >> when_what_where_how_why; n = when_what_where_how_why.size(); for (int i = 1; i <= n; i++) a[i] = when_what_where_how_why[i - 1]; powbase[0] = invpowbase[0] = 1; for (int i = 1; i <= n; i++) { powbase[i] = base * powbase[i - 1] % modulo; if (i >= 2) invpowbase[i] = invpowbase[1] * invpowbase[i - 1] % modulo; else invpowbase[i] = lgpow(base,modulo - 2); } for (int i = 1; i <= n; i++) h[i] = (h[i - 1] + (a[i] - 'a' + 1) * powbase[i - 1]) % modulo; for (int i = n; i >= 1; i--) revh[i] = (revh[i + 1] + (a[i] - 'a' + 1) * powbase[n - i]) % modulo; for (int i = 1; i <= n; i++) { int st = 0,pas = 1 << 18; while (pas != 0) { if (pal(i - st - pas,i + st + pas)) st += pas; pas /= 2; } ev[i].push_back(i); ev[i + st + 1].push_back(-i); } for (int i = 1; i < n; i++) { int st = 0,pas = 1 << 18; while (pas != 0) { if (pal(i - st - pas + 1,i + st + pas)) st += pas; pas /= 2; } if (st == 0) continue; double lol = (double)i + 0.5d; ev[i + 1].push_back(lol); ev[i + st + 1].push_back(-lol); } multiset<double> ms; for (int i = 1; i <= n; i++) { for (auto it : ev[i]) { double vl = it; if (vl > 0) ms.insert(vl); else ms.erase(ms.find(-vl)); } double hmm = *ms.begin(); double d = (double)i - hmm; d *= (2.0d); lft[i] = i - d; if (ms.size() == 1) continue; ms.erase(ms.find(hmm)); double hmm2 = *ms.begin(); d = (double)i - hmm2; d *= (2.0d); l2[i] = i - d; ms.insert(hmm); } for (int i = 1; i <= n; i++) { int cur = get_hash(lft[i],i); if (!id[cur]) id[cur] = ++cnt_id; what[id[cur]] = {lft[i],i}; f[id[cur]]++; } vector<pair<int,int>> ord; for (int i = 1; i <= cnt_id; i++) ord.push_back({what[i].second - what[i].first + 1,i}); sort(ord.begin(),ord.end()); reverse(ord.begin(),ord.end()); int ans = 0; for (auto it : ord) { ans = max(ans,it.first * f[it.second]); if (it.first == 1) continue; int h2 = get_hash(l2[it.second],it.second); int id2 = id[h2]; f[id2] += f[it.second]; } cout << ans; return 0; } /** Maxim N palindroame distincte etc Voi vrea pentru fiecare sa retin frecventa lui Pentru asta, merg cu i de la 1 la N, iau cel mai mare palindrom care se termina pe i Ii gasesc id-ul (ori un id nou ori ceva id vechi fiindca deja exista) Acum, vreau sa adaug 1 pe toate palindroamele care se termina pe i <=> toate sufixele palindrom ale lui pal[id] <=> toate prefixele palindrom Dau f[id]++ La final, vreau sa propag f-urile pe toate prefixele palindrom Foarte simplu, iau palindroamele descrescator dupa lungime, pentru asta o sa am f[id] updatat si il consider la raspuns Fie pal[id'] = prefixul maxim palindrom al lui pal[id] (daca pal[id] are lungime 1, ma opresc) Dau f[id'] += f[id] and we move on **/
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