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
[<}){}
2
{()}[]
0
]]
Impossible
题意：输入一行括号序列，左边括号可以和右括号抵消。右括号可以转换为其他类型右括号进行成对抵消。输出需要转换的个数。如果无法抵消输出"Impossible"
1 #include <iostream> 2 #include <stack> 3 using namespace std; 4 5 int main(){ 6 string input; 7 while(cin >> input){ 8 int count = 0; 9 stack<char> s; 10 s.push(input[0]); //[ 11 for(int i = 1;i < input.length();i++){//<}){} 12 if(s.size() > 0 && (s.top() == ‘[‘ || s.top() == ‘<‘ || s.top() == ‘(‘ || s.top() == ‘{‘)){ 13 if(input[i] == ‘]‘ || input[i] == ‘>‘ || input[i] == ‘)‘ || input[i] == ‘}‘){ 14 if(!((s.top() == ‘[‘ && input[i] == ‘]‘) || 15 (s.top() == ‘<‘ && input[i] == ‘>‘) || 16 (s.top() == ‘(‘ && input[i] == ‘)‘) || 17 (s.top() == ‘{‘ && input[i] == ‘}‘))){ 18 count++; 19 } 20 s.pop(); 21 }else{ 22 s.push(input[i]); 23 } 24 }else{ 25 s.push(input[i]); 26 } 27 } 28 if(s.size() != 0){ 29 cout << "Impossible" << endl; 30 }else{ 31 cout << count << endl; 32 } 33 } 34 }
Replace To Make Regular Bracket Sequence
原文：https://www.cnblogs.com/jxust-jiege666/p/10778764.html