题解

题意：

括号匹配，给一段括号，可以选择一段区间[l,r][l,r][l,r]，交换其内部括号的位置使得这一段区间的括号是匹配的，代价是区间长度，问要使整个区间都是匹配的最小代价是多少

官方题解：

遇到 (((，就+1，遇到 )))，就 -1，找离-1最近的0的位置，就是这一段区间最小的代价

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;
int a[N];
int n, m, K;
string s;int main() {ios::sync_with_stdio(0);cin >> n;cin >> s;for (int i = 1; i <= n; ++i) {a[i] = a[i - 1];if (s[i - 1] == '(') a[i]++;else a[i]--;}if(a[n]) return puts("-1"),0;int i = 1, ans = 0;while (i <= n) {if (a[i] != -1) {i++;continue;}int j = i;while (j <= n && a[j]) {j++;}ans += j - i + 1;i = j;}cout <<ans << endl;return 0;
}

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