CF思维联系–CodeForces -224C

ACM思维题训练集合
A bracket sequence is a string, containing only characters “(”, “)”, “[” and “]”.

A correct bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters “1” and “+” between the original characters of the sequence. For example, bracket sequences “()[]”, “([])” are correct (the resulting expressions are: “(1)+[1]”, “([1+1]+1)”), and “](” and “[” are not. The empty string is a correct bracket sequence by definition.

A substring s[l… r] (1 ≤ l ≤ r ≤ |s|) of string s = s1s2… s|s| (where |s| is the length of string s) is the string slsl + 1… sr. The empty string is a substring of any string by definition.

You are given a bracket sequence, not necessarily correct. Find its substring which is a correct bracket sequence and contains as many opening square brackets «[» as possible.

Input
The first and the only line contains the bracket sequence as a string, consisting only of characters “(”, “)”, “[” and “]”. It is guaranteed that the string is non-empty and its length doesn’t exceed 105 characters.

Output
In the first line print a single integer — the number of brackets «[» in the required bracket sequence. In the second line print the optimal sequence. If there are more than one optimal solutions print any of them.
Examples

Input
([])
Output
1
([])
Input
(((
Output
0
括号是就近匹配的，所以可以用栈来模拟，所以可以将括号压栈，匹配后出栈，最后栈底剩余的就是不能出栈的就是不能匹配的，一般的方法是找到这些括号但是太费劲了，我们同时建立一个栈，同时入栈，出栈，存括号的下标，那么在出栈操作之后，第一个stack就只剩下不匹配的括号，第二个stack就只剩下不匹配的括号的下标。
下标将括号数组分成了好几段，枚举每一段的左中括号的数量即可，比较最大值更新左右段点即可

#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
    char ch = getchar();
    x = 0;
    int f = 1;
    while (ch < '0' || ch > '9')
        f = (ch == '-' ? -1 : f), ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = x * 10 + ch - '0', ch = getchar();
    x *f;
}
#define wi(n) printf("%d ", n)
#define wl(n) printf("%lld ", n)
#define P puts(" ")
typedef long long ll;
#define MOD 1000000007
#define mp(a, b) make_pair(a, b)
//---------------https://lunatic.blog.csdn.net/-------------------//
const int N = 1e5 + 5;
char a[N];
stack<char> m;
stack<int> n;
vector<int> x;
int main()
{
    scanf("%s", a);
    int ln = strlen(a);
    int cnt = 0;
    //cout<<ln<<endl;
    for (int i = 0; i < ln; i++)
    {

        if (!m.size() || a[i] == '(' || a[i] == '[')
        {
            m.push(a[i]);
            n.push(i);
            continue;
        }
        else if (a[i] == ')' && m.top() == '(')
        {

            n.pop();
            m.pop();
            //cout << 1 << endl;
        }
        else if (a[i] == ']' && m.top() == '[')
        {
            m.pop();
            n.pop();
            cnt++;
            // cout << 2 << endl;
        }
        else
        {
            m.push(a[i]);
            n.push(i);
        }
    }
    // cout<<1<<endl;
    if (m.empty())
    {
        wi(cnt);
        P;
        printf("%s
", a);
    }
    else
    {
        x.push_back(ln);
        while (!n.empty())
        {
            x.push_back(n.top());
            n.pop();
        }
        x.push_back(-1);
        int ml, mr, maxi = 0;
        cnt = 0;
        for (int i = x.size() - 1; i > 0; i--)
        {
            maxi = 0;
          //  cout << x[i] + 1 << " " << x[i - 1] - 1 << endl;

            for (int j = x[i] + 1; j < x[i - 1]; j++)
            {
                if (a[j] == '[')
                    maxi++;
            }
            if (maxi > cnt)
            {
                cnt = maxi;
                ml = x[i] + 1;
                mr = x[i - 1] - 1;
            }
        }
        wi(cnt);
        P;
        if (cnt == 0)
            return 0;

        for (int i = ml; i <= mr; i++)
        {
            putchar(a[i]);
        }
        P;
    }
    //P;
}