栈
Description
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 Input
[<}){}
2
{()}[]
0
]]
1 #include<stdio.h> 2 #include<stack> 3 #include<queue> 4 #include<string.h> 5 using namespace std; 6 int main() 7 { char s[1000000+10]; 8 while(scanf("%s",&s)!=EOF) 9 { 10 stack<char>sta; 11 int i,j,k,l,ans=1; 12 l=strlen(s); 13 14 for(k=0,i=0;i<l;i++) 15 { 16 if(sta.empty()&&(s[i]==']'||s[i]=='}'||s[i]=='>'||s[i]==')')) 17 { ans=0; 18 break; 19 } 20 else if(s[i]=='<'||s[i]=='{'||s[i]=='['||s[i]=='(') 21 { sta.push(s[i]); 22 continue; 23 } 24 else 25 { if(s[i]-sta.top()==2||s[i]-sta.top()==1) 26 sta.pop(); 27 else 28 { sta.pop(); 29 k++; 30 } 31 32 } 33 } 34 if(ans==0||!sta.empty()) 35 printf("Impossible\n"); 36 else 37 printf("%d\n",k); 38 } 39 return 0; 40 }