186 words
1 minutes
[BOJ] 괄호 제거
| TimeLimit | MemoryLimit | Condition | TAG |
|---|---|---|---|
| 1s | 128MB | (1 <= 괄호 개수 <= 10) | Implementation |
오랜만에 문제를 풀기 때문에 가벼운 문제 입니다. ㅠ
정답 코드
#include <iostream>
#include <queue>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <sstream>
#include <iomanip>
#include <string>
#include <numeric>
#include <unordered_set>
#include <unordered_map>
#include <climits>
#include <math.h>
using namespace std;
using ll = long long;
using ld = long double;
ll MOD = 1'000'000'000;
set<string> results;
string input;
set<int> visited;
void solution(int left){
if(left >= input.size()) return;
if(input[left] == '('){
int right_idx = input.size()-1;
visited.insert(left);
int cnt = 1;
for(int i = left+1; i < input.size(); i++){
if(input[i] == ')') cnt--;
else if(input[i] == '(') cnt++;
if(cnt == 0){
right_idx = i;
break;
}
}
visited.insert(right_idx);
solution(left+1);
string result = "";
for(int i = 0; i < input.size(); i++){
if(visited.find(i) == visited.end())
result += input[i];
}
results.insert(result);
visited.erase(left);
visited.erase(right_idx);
solution(left+1);
}
else{
solution(left+1);
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> input;
solution(0);
for(string result : results) cout << result << "\n";
return 0;
}

