제출 #874975

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
874975		_fractal	괄호 문자열 (CEOI16_match)	C++14	100 / 100	6 ms	12124 KiB

match

#include <bits/stdc++.h>
using namespace std;

#define F first
#define S second 
#define mp make_pair
#define pb push_back
#define pf push_front
#define ppb pop_back
#define ppf pop_front
#define speed ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define sz(x) (int)x.size()
#define len(x) (int)strlen(x)
#define all(x) x.begin(), x.end()
#define debug cerr << "OK\n";
#define ub upper_bound
#define lb lower_bound 
#define make_unique(x) sort(all(x)), x.erase(unique(all(x)), x.end())

mt19937 bruh(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rofl(chrono::steady_clock::now().time_since_epoch().count());

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef set<int> si;
typedef set<ll> sll;
typedef set<pii> spii;
typedef set<pll> spll;
typedef multiset <int> msi;
typedef multiset <ll> msll;
typedef map <int, int> mi;
typedef map <ll, ll> mll;

const int N = 1e5 + 2;
const int M = 1e5;
const int inf = 2e9 + 3;
const ll INF = 1e18;
const ld pi2 = 2.0 * 3.141592653589793;
const ld pi = 3.141592653589793;
const ld eps = 1e-12;

const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, -1, 1};

string s, t;
int n;
int dp[N][26];

void solve(int l, int r) {
    if (l > r) return;
    int ch = (s[l] - 'a');
    if (dp[r][ch] <= l) return;
    int p = dp[r][ch];
    t[l] = '(';
    t[p] = ')';
    solve(l + 1, p - 1);
    solve(p + 1, r);
}

int main() {
    speed;
    cin >> s;
    n = sz(s);
    t = string(n, '0');
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < 26; ++j) {
            dp[i][j] = -1;
        }
        int ch = (s[i] - 'a');
        int p = (i > 0 ? dp[i - 1][ch] : -1);
        for (int j = 0; j < 26; ++j) {
            if (p > 0) {
                if (s[p - 1] - 'a' == j)
                    dp[i][j] = max(dp[i][j], p - 1);
                else 
                    dp[i][j] = max(dp[i][j], dp[p - 1][j]);
            }
        }
        dp[i][ch] = i;
    }
    solve(0, n - 1);
    bool F = 1;
    for (auto c : t) if (c == '0') F = 0;
    if (!F) cout << -1 << '\n';
    else cout << t << '\n';
    
}

• Subtask 1: 10 / 10
• Subtask 2: 27 / 27
• Subtask 3: 63 / 63