【CodeForces】223A - Bracket Sequence（栈 & 模拟）

最新推荐文章于 2020-07-06 22:16:34 发布

wyg1997

最新推荐文章于 2020-07-06 22:16:34 发布

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分类专栏：模拟栈和队列

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栈和队列同时被 2 个专栏收录

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探讨括号匹配问题中的最优子序列寻找方法，通过栈结构实现有效的匹配过程，并找到包含最多方括号的正确括号序列。

点击打开题目

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

这道题做起来好烦啊，说一下思路吧：

像平时括号配对问题一样，能配对的弹出，否则就压入栈（下标），最后得到的栈两两数中间是能配对的（不包括开头结尾），然后依次取数算次数就行了。

代码如下：

#include <cstdio>
#include <map>
#include <stack>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
	map<char,int> m;
	m['('] = 1;
	m[')'] = -1;
	m['['] = 2;
	m[']'] = -2;
	char str[100000+22];
	int l;
	int ans;
	while (~scanf ("%s",str))
	{
		l = strlen(str);
		stack<int> s;
		ans = 0;
		for (int i = 0 ; i < l ; i++)
		{
			if (m[str[i]] == 1 || m[str[i]] == 2)
				s.push(i);
			else if (s.empty())
				s.push(i);
			else if (m[str[i]] + m[str[s.top()]] == 0)		//可以配对，弹出
				s.pop();
			else		//不能配对，压入 
				s.push(i);
		}
		
		if (s.empty())		//全部完成匹配 
		{
			for (int i = 0 ; i < l ; i++)
				if (m[str[i]] == 2)
					ans++;
			printf ("%d\n%s\n",ans,str);
		}
		else
		{
			int st,endd;		
			int ans_st,ans_endd;		//满足条件的开始于结束位置
			int ant;
			s.push(l);
			while (s.size() != 1)
			{
				endd = s.top() - 1;
				s.pop();
				st = s.top() + 1;
				ant = 0;
				for (int i = st ; i <= endd ; i++)
					if (m[str[i]] == 2)
						ant++;
				if (ant > ans)
				{
					ans = ant;
					ans_st = st;
					ans_endd = endd;
				}
			}
			st = 0;
			endd = s.top() - 1;
			ant = 0;
			for (int i = st ; i <= endd ; i++)
				if (m[str[i]] == 2)
					ant++;
			if (ant > ans)
			{
				ans = ant;
				ans_st = st;
				ans_endd = endd;
			}
			if (ans)
			{
				printf ("%d\n",ans);
				for (int i = ans_st ; i <= ans_endd ; i++)
					printf ("%c",str[i]);
				printf ("\n");
			}
			else
				printf ("0\n\n");
		}
	}
	return 0;
}