括号配对 栈

 IIITime Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64uSubmit StatusDescriptionYou 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.
InputThe 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.
OutputIf it's impossible to get RBS from s print Impossible.
Otherwise print the least number of replaces needed to get RBS from s.
Sample InputInput

[<}){}

Output

2

Input

{()}[]

Output

0

Input

]]

Output

Impossible

题意：现在有一段字符串，称'<'、'{'、'['、'('为开类型，反之为闭类型，只有开类型可以转换，现在问最少得转换多少次才使得整个字符串中括号配对，如过不行就输出“Impossible”.

思路：用栈模拟，遍历字符串数组，只能放进开类型，当遇到闭类型时，如果栈顶的字符是和它匹配的就不用转换，否则需要转换；当遇到闭类型时，如果栈为空的话，这个时候就是“Impossible”，因为闭类型不能转换成开类型。最后如果栈不为空也是“Impossible”。当时理解错题意了。

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <stack>
using namespace std;
char s[1000000];
char p(char a)
{
if(a=='{') return '}';
if(a=='<') return '>';
if(a=='[') return ']';
if(a=='(') return ')';
}
int main()
{

while(~scanf("%s", s))
{
stack<char> sta;
int len = strlen(s);
int t = 0;
for(int i = 0;i < len;i++)
{
if(s[i]=='<'||s[i]=='{'||s[i]=='['||s[i]=='(')
sta.push(s[i]);
else if(!sta.empty())
{
if(p(sta.top()) != s[i])
t++;
sta.pop();
}
else
{
printf("Impossible\n");
return 0;
}
}
if(!sta.empty()) printf("Impossible\n");
else printf("%d\n", t);
}
}