oj.uz

답안 #84101

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
84101 2018-11-12T21:08:04 Z radoslav11 괄호 문자열 (CEOI16_match) C++14
100 / 100
 25 ms 12940 KB 
/*
   We will solve the problem recursively. It's intuitive that we want to match the bracket at position 0 to the rightmost possible position. Lets have T = S[p + 1; N - 1].
   If both S and T have at least one possible solving pattern then S[1; p - 1] also has a solving pattern and we can match 0 with p. So we will find the largest such p with S[p] == S[0].
   Then we will solve the problem recursively on S[1, p - 1] and S[p + 1, N - 1]. If we can find p fast, the complexity of the recursion is O(N).
   */

#include <bits/stdc++.h>
#define endl '\n'

//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")

#define SZ(x) ((int)x.size())
#define ALL(V) V.begin(), V.end()

using namespace std;
template<class T, class T2> inline int chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline int chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const int MAXN = (int)1e5 + 42;
const int SIGMA = 26;

string s;

void read()
{
	cin >> s;
}

char answer[MAXN];
int last[MAXN][SIGMA];

void rec(int l, int r)
{
	if(r < l)
		return;

	int mid = last[r][s[l] - 'a'];

	answer[l] = '(';
	answer[mid] = ')';

	rec(l + 1, mid - 1);
	rec(mid + 1, r);
}

void solve()
{
	stack<int> st;
	for(int i = 0; i < SZ(s); i++)
		if(!st.empty() && s[st.top()] == s[i]) st.pop();
		else st.push(i);

	if(!st.empty())
	{
		cout << -1 << endl;
		return;
	}

	memset(last, -1, sizeof(last));

	for(int i = 0; i < SZ(s); i++)
	{
		last[i][s[i] - 'a'] = i;
		if(i >= 1)
		{
			int pos = last[i - 1][s[i] - 'a'];
			for(int c = 0; c < SIGMA; c++)
				if(pos > 0 && c != (s[i] - 'a'))
					last[i][c] = last[pos - 1][c];
		}
	}

	rec(0, SZ(s) - 1);

	for(int i = 0; i < SZ(s); i++)
		cout << answer[i];
	cout << endl;
}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	read();
	solve();
	return 0;
}

Subtask #1
10.0 / 10.0

# 결과 실행 시간 메모리 Grader output
1 Correct 10 ms 10488 KB Output is correct
2 Correct 10 ms 10612 KB Output is correct
3 Correct 2 ms 10612 KB Output is correct

Subtask #2
27.0 / 27.0

# 결과 실행 시간 메모리 Grader output
1 Correct 10 ms 10488 KB Output is correct
2 Correct 10 ms 10612 KB Output is correct
3 Correct 2 ms 10612 KB Output is correct
4 Correct 10 ms 10612 KB Output is correct
5 Correct 10 ms 10612 KB Output is correct
6 Correct 10 ms 10644 KB Output is correct
7 Correct 10 ms 10776 KB Output is correct

Subtask #3
63.0 / 63.0

# 결과 실행 시간 메모리 Grader output
1 Correct 10 ms 10488 KB Output is correct
2 Correct 10 ms 10612 KB Output is correct
3 Correct 2 ms 10612 KB Output is correct
4 Correct 10 ms 10612 KB Output is correct
5 Correct 10 ms 10612 KB Output is correct
6 Correct 10 ms 10644 KB Output is correct
7 Correct 10 ms 10776 KB Output is correct
8 Correct 11 ms 10784 KB Output is correct
9 Correct 11 ms 10840 KB Output is correct
10 Correct 11 ms 10912 KB Output is correct
11 Correct 11 ms 10932 KB Output is correct
12 Correct 16 ms 11596 KB Output is correct
13 Correct 16 ms 11784 KB Output is correct
14 Correct 17 ms 12292 KB Output is correct
15 Correct 18 ms 12496 KB Output is correct
16 Correct 25 ms 12576 KB Output is correct
17 Correct 20 ms 12780 KB Output is correct
18 Correct 19 ms 12780 KB Output is correct
19 Correct 19 ms 12780 KB Output is correct
20 Correct 17 ms 12780 KB Output is correct
21 Correct 21 ms 12940 KB Output is correct