Replace To Make Regular Bracket Sequence CodeForces

本文介绍了一种解决括号匹配问题的方法，通过自定义堆栈结构实现括号的有效配对，进而确定最少替换次数使字符串成为合法的括号序列。

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Replace To Make Regular Bracket Sequence - CodeForces 612C - Virtual Judgehttps://vjudge.net/problem/CodeForces-612C

You are given string s consists of opening and closing brackets of four kinds <>, {}, [], (). There are two types of brackets: opening and closing. You can replace any bracket by another of the same type. For example, you can replace < by the bracket {, but you can't replace it by ) or >.

The following definition of a regular bracket sequence is well-known, so you can be familiar with it.

Let's define a regular bracket sequence (RBS). Empty string is RBS. Let s1 and s2 be a RBS then the strings <s1>s2, {s1}s2, [s1]s2, (s1)s2 are also RBS.

For example the string "[[(){}]<>]" is RBS, but the strings "[)()" and "][()()" are not.

Determine the least number of replaces to make the string s RBS.

Input

The only line contains a non empty string s, consisting of only opening and closing brackets of four kinds. The length of s does not exceed 106.

Output

If it's impossible to get RBS from s print Impossible.

Otherwise print the least number of replaces needed to get RBS from s.

Examples

Input

[<}){}

Output

Input

{()}[]

Output

Input

]]

Output

Impossible

#include<iostream>
using namespace std;
#include<string>
#include<algorithm>
#pragma warning (disable:4996)
#include <climits>
#include <vector>
typedef struct a
{
	char zuo[1000000];
	int top = -1;
}node;
int main() {
	node a;
	char temp;
	int ans = 0;
	while ((temp = getchar()) != '\n') {
		if (temp == '[' || temp == '<' || temp == '{' || temp == '(') {
			a.zuo[++a.top] = temp;
		}
		else if((temp==']'||temp=='>'||temp=='}' || temp==')') && a.top!=-1)
		{
			if (('{' == a.zuo[a.top] && temp == '}' || a.zuo[a.top] == '[' && temp == ']' || a.zuo[a.top] == '<' && temp == '>' || a.zuo[a.top] == '(' && temp == ')') )
				a.zuo[a.top--] = 0;
			else  {
				ans++;
				a.zuo[a.top--] = 0;
			}
		}
		else if((temp == ']' || temp == '>' || temp == '}' || temp == ')') && a.top == -1)
		{
			cout << "Impossible";
			return 0;
		}
	}
	if (a.top != -1)cout << "Impossible";
	else cout << ans;
	return 0;
}

这题因为我第一次自己创建堆栈,所以思路没有太大问题,反而是编译的时候错误不断,我真是吐了,一直给我越界,然后下面这个就代表你数组越界了

Run-Time Check Failure #2 - Stack around the variable 'a' was corrupted.

我用stlstack重写了一遍,会更清晰点

#include<iostream>
using namespace std;
#include<string>
#include<algorithm>
#pragma warning (disable:4996)
#include <climits>
#include <vector>
#include<stack>
int main() {
	stack<char> a;
	char temp;
	int ans = 0;
	while ((temp = getchar()) != '\n') {
		if (temp == '[' || temp == '{' || temp == '<' || temp == '(')
			a.push(temp);
		else if (temp == ']' || temp == '}' || temp == ')' || temp == '>') {
			if (a.empty()) {
				cout << "Impossible";
				return 0;
			}
			else
			{
				char zuo = a.top();
				a.pop();
				if (zuo == '[' && temp == ']' || zuo == '{' && temp == '}' || zuo == '(' && temp == ')' || zuo == '<' && temp == '>') {
					continue;
				}
				else
				{
					ans++;
				}
			}
		}
	}
	if (a.empty())cout << ans;
	else cout << "Impossible";
	return 0;
}

然后注意下,我重写的时候忘记了最后还有一种就是左边括号太多的情况,所以最后加了个empty在判断下