Codeforces 612C Replace To Make Regular Bracket Sequence 【stack】

C. Replace To Make Regular Bracket Sequence

time limit per test
 1 second

memory limit per test
 256 megabytes

input
 standard input

output
 standard output

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.

Sample test(s)

input

[<}){}

output

2

input

{()}[]

output

0

input

]]

output

Impossible

题意：给定一个只有'{' '[' '(' '<'开字符和'}' ']' ')' '>'闭字符的字符串，你可以将开字符互相转化，闭字符互相转化，要求得到一个括号匹配的串。若可以输出最少转化次数，反之输出Impossible。

记录左右括号的数量，若不相等，则说明没有#include<cstdio>
#include<stack>
#include<cstring>
using namespace std;
int main()
{
stack<char> s;
char c[1000010];
int i,l;
while(scanf("%s",c)!=EOF)
{
int ans=0,lift=0;;
l=strlen(c);
for(i=0;i<l;i++)
{
if(c[i]=='{'||c[i]=='<'||c[i]=='['||c[i]=='(')
{
s.push(c[i]);
lift++;
}
else
{
if(c[i]=='}')
{
if(s.empty())
break;
else if(s.top()=='{')
s.pop();
else
{
ans++; s.pop();
}
}
else if(c[i]==']')
{
if(s.empty())
break;
else if(s.top()=='[')
s.pop();
else
{
ans++; s.pop();
}
}
if(c[i]=='>')
{
if(s.empty())
break;
else if(s.top()=='<')
s.pop();
else
{
ans++; s.pop();
}
}
if(c[i]==')')
{
if(s.empty())
break;
else if(s.top()=='(')
s.pop();
else
{
ans++;
s.pop();
}
}

}
}
if(double(l)/2!=double(lift))
printf("Impossible\n");
else if(i==l)
printf("%d\n",ans);
else printf("Impossible\n");
}
return 0;
}

完成配对。