제출 #612290

#제출 시각아이디문제언어결과실행 시간메모리
612290Bungmint회문 (APIO14_palindrome)C++17
0 / 100
1090 ms988 KiB
// Copyright © 2022 Youngmin Park. All rights reserved. //#pragma GCC optimize("O3") //#pragma GCC target("avx2") #include <bits/stdc++.h> using namespace std; #pragma region TEMPLATE using ll = long long; using vi = vector<int>; using pii = pair<int, int>; using vpi = vector<pii>; using pll = pair<ll, ll>; using vl = vector<ll>; using vpl = vector<pll>; using ld = long double; template <typename T, size_t SZ> using ar = array<T, SZ>; template <typename T> using pqg = priority_queue<T, vector<T>, greater<T>>; #define all(v) (v).begin(), (v).end() #define pb push_back #define sz(x) (int)(x).size() #define fi first #define se second #define lb lower_bound #define ub upper_bound constexpr int INF = 1e9; constexpr ll LINF = 1e18; const ld PI = acos((ld)-1.0); constexpr int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1}; mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); template <typename T> constexpr bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; } template <typename T> constexpr bool ckmax(T &a, const T &b) { return b > a ? a = b, 1 : 0; } ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); } // divide a by b rounded up ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); } // divide a by b rounded down #ifdef LOCAL #include "miscellaneous/debug.h" #else #define dbg(...) 42 #endif inline namespace RecursiveLambda { template <typename Fun> struct y_combinator_result { Fun fun_; template <typename T> explicit constexpr y_combinator_result(T &&fun) : fun_(forward<T>(fun)) {} template <typename... Args> constexpr decltype(auto) operator()(Args &&...args) const { return fun_(ref(*this), forward<Args>(args)...); } }; template <typename Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<decay_t<Fun>>(forward<Fun>(fun)); } }; #pragma endregion TEMPLATE /** * Description: Trie-like structure storing distinct palindromes. Also called eertree * Edge from u to v with character c: S_v = "c" + S_u + "c" * Suffix links: Links to the longest nontrivial palindromic suffix * Root 1 = Length -1, Root 2 = Length 0 * Source: http://adilet.org/blog/palindromic-tree/ * Verification: * Time complexity: O(|S|) * Memory complexity: O(\sigma |S|) but can be reduced by not storing all edges * Check adamant's comment - https://codeforces.com/blog/entry/13959?#comment-313090 */ struct Node { int to[26]; int len; int suf_link; }; const int MAXN = 3e5+10; Node tree[MAXN*26]; string s; int ind, last_suf; bool add_char(int pos) { int c = s[pos] - 'a'; int cur = last_suf, cur_len = 0; while (1) { cur_len = tree[last_suf].len; if (pos-1-cur_len >= 0 && s[pos-1-cur_len] == s[pos]) { break; } cur = tree[cur].suf_link; } if (tree[cur].to[c]) { last_suf = tree[cur].to[c]; return false; } ind++; tree[cur].to[c] = ind; tree[ind].len = 2 + tree[cur].len; if (tree[ind].len == 1) { tree[ind].suf_link = 2; return true; } while (1) { cur = tree[cur].suf_link; cur_len = tree[cur].len; if (pos-1-cur_len >= 0 && s[pos-1-cur_len] == s[pos]) { tree[ind].suf_link = tree[cur].to[c]; return true; } } } void init_tree() { tree[1].len = 0; tree[2].len = -1; tree[1].suf_link = tree[2].suf_link = 1; ind = last_suf = 2; } void solve() { cin >> s; int n = sz(s); init_tree(); for (int i = 0; i < n; i++) { add_char(i); } } int main() { cin.tie(0)->sync_with_stdio(0); cin.exceptions(cin.failbit); int testcase = 1; // cin >> testcase; while (testcase--) { solve(); } #ifdef LOCAL cerr << "Time elapsed: " << 1.0 * (double)clock() / CLOCKS_PER_SEC << " s.\n"; #endif }

컴파일 시 표준 에러 (stderr) 메시지

palindrome.cpp:7: warning: ignoring '#pragma region TEMPLATE' [-Wunknown-pragmas]
    7 | #pragma region TEMPLATE
      | 
palindrome.cpp:65: warning: ignoring '#pragma endregion TEMPLATE' [-Wunknown-pragmas]
   65 | #pragma endregion TEMPLATE
      |
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