CodeForces - 224C. Bracket Sequence (栈模拟）简单做法

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;
const int maxn=100005;
int b[maxn]={0};
stack<int>demo;
stack<int>de;
vector<int>ans;
string a;
int main()
{while(!demo.empty())demo.pop();while(!de.empty())de.pop();ans.clear(); //初始化操作cin>>a;int n=a.size();for(int i=0;i<a.size();i++){if(a[i]=='(') b[i]=-2;else  if(a[i]==')') b[i]=2;else if(a[i]=='[') b[i]=-1;else b[i]=1;}for(int i=0;i<n;i++){if(demo.empty()||b[i]==-1||b[i]==-2){demo.push(b[i]);de.push(i);}else if(b[i]+demo.top()==0){demo.pop();de.pop();}else{demo.push(a[i]);de.push(i);}}if(demo.empty()){int res=0;for(int i=0;i<a.size();i++) if(a[i]=='[') res++;cout<<res<<endl<<a<<endl;return 0;}while(!de.empty()){ans.push_back(de.top());de.pop();}ans.push_back(n);//补足区间sort(ans.begin(),ans.end());int l,r=-1, /*补足区间*/ml,mr,res=0,maxi=0;for(int i=0;i<ans.size();i++){l=r+1;r=ans[i];res=0;for(int i=l;i<r;i++){if(a[i]=='[') res++;}if(res>maxi){ml=l;mr=r-1;maxi=res;}}if(maxi==0){cout<<0<<endl;return 0;}cout<<maxi<<endl;for(int i=ml;i<=mr;i++)  cout<<a[i];cout<<endl;
}

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