結果

問題 No.2204 Palindrome Splitting (No Rearrangement ver.)
ユーザー tokusakurai
提出日時 2023-02-03 21:44:25
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 42 ms / 2,000 ms
コード長 4,318 bytes
コンパイル時間 1,706 ms
コンパイル使用メモリ 197,456 KB
最終ジャッジ日時 2025-02-10 09:00:40
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 33
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define per(i, n) for (int i = (n)-1; i >= 0; i--)
#define rep2(i, l, r) for (int i = (l); i < (r); i++)
#define per2(i, l, r) for (int i = (r)-1; i >= (l); i--)
#define each(e, v) for (auto &e : v)
#define MM << " " <<
#define pb push_back
#define eb emplace_back
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T>;
template <typename T>
bool chmax(T &x, const T &y) {
return (x < y) ? (x = y, true) : false;
}
template <typename T>
bool chmin(T &x, const T &y) {
return (x > y) ? (x = y, true) : false;
}
template <typename T>
int flg(T x, int i) {
return (x >> i) & 1;
}
template <typename T>
void print(const vector<T> &v, T x = 0) {
int n = v.size();
for (int i = 0; i < n; i++) cout << v[i] + x << (i == n - 1 ? '\n' : ' ');
if (v.empty()) cout << '\n';
}
template <typename T>
void printn(const vector<T> &v, T x = 0) {
int n = v.size();
for (int i = 0; i < n; i++) cout << v[i] + x << '\n';
}
template <typename T>
int lb(const vector<T> &v, T x) {
return lower_bound(begin(v), end(v), x) - begin(v);
}
template <typename T>
int ub(const vector<T> &v, T x) {
return upper_bound(begin(v), end(v), x) - begin(v);
}
template <typename T>
void rearrange(vector<T> &v) {
sort(begin(v), end(v));
v.erase(unique(begin(v), end(v)), end(v));
}
template <typename T>
vector<int> id_sort(const vector<T> &v, bool greater = false) {
int n = v.size();
vector<int> ret(n);
iota(begin(ret), end(ret), 0);
sort(begin(ret), end(ret), [&](int i, int j) { return greater ? v[i] > v[j] : v[i] < v[j]; });
return ret;
}
template <typename S, typename T>
pair<S, T> operator+(const pair<S, T> &p, const pair<S, T> &q) {
return make_pair(p.first + q.first, p.second + q.second);
}
template <typename S, typename T>
pair<S, T> operator-(const pair<S, T> &p, const pair<S, T> &q) {
return make_pair(p.first - q.first, p.second - q.second);
}
template <typename S, typename T>
istream &operator>>(istream &is, pair<S, T> &p) {
S a;
T b;
is >> a >> b;
p = make_pair(a, b);
return is;
}
template <typename S, typename T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
return os << p.first << ' ' << p.second;
}
struct io_setup {
io_setup() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout << fixed << setprecision(15);
}
} io_setup;
const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
// const int MOD = 1000000007;
const int MOD = 998244353;
struct Union_Find_Tree {
vector<int> data;
const int n;
int cnt;
Union_Find_Tree(int n) : data(n, -1), n(n), cnt(n) {}
int root(int x) {
if (data[x] < 0) return x;
return data[x] = root(data[x]);
}
int operator[](int i) { return root(i); }
bool unite(int x, int y) {
x = root(x), y = root(y);
if (x == y) return false;
if (data[x] > data[y]) swap(x, y);
data[x] += data[y], data[y] = x;
cnt--;
return true;
}
int size(int x) { return -data[root(x)]; }
int count() { return cnt; };
bool same(int x, int y) { return root(x) == root(y); }
void clear() {
cnt = n;
fill(begin(data), end(data), -1);
}
};
int main() {
string S;
cin >> S;
int N = sz(S);
vector<int> dp(N + 1, -inf);
dp[0] = inf;
vector<bool> ok(N, true), ok2(N - 1, true);
rep(i, N) {
rep(j, i + 1) {
if (ok[j]) {
if (2 * j - i < 0 || S[2 * j - i] != S[i]) ok[j] = false;
}
if (ok[j]) chmax(dp[i + 1], min(dp[2 * j - i], 2 * (i - j) + 1));
}
rep(j, i) {
if (ok2[j]) {
if (2 * j - i + 1 < 0 || S[2 * j - i + 1] != S[i]) ok2[j] = false;
}
if (ok2[j]) chmax(dp[i + 1], min(dp[2 * j - i + 1], 2 * (i - j)));
}
}
cout << dp[N] << '\n';
}
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