CodeForces 612C Replace To Make Regular Bracket Sequence (栈)
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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
]]
Impossible
#include<stdio.h>#include<string.h>#include<stack>#include<algorithm>using namespace std;char str[1100000];int main(){int i,j,l,m,n,k;while(scanf("%s",str)!=EOF){ stack<char>s; int ans=0; int flag=0;int len=strlen(str);k=l=0;for(i=0;i<len;i++){if(str[i]=='('||str[i]=='<'||str[i]=='{'||str[i]=='['){s.push(str[i]);k++;}else if(str[i]==')'){l++;if(s.empty()){flag=1;break;}else {if(s.top()=='(')s.pop();else{ans++;s.pop();}}}else if(str[i]=='>'){l++;if(s.empty()){flag=1;break;}else{if(s.top()=='<')s.pop();else{ans++;s.pop();}}}else if(str[i]=='}'){l++;if(s.empty()){flag=1;break;}else{if(s.top()=='{')s.pop();else{ans++;s.pop();}}}else if(str[i]==']'){l++;if(s.empty()){flag=1;break;}else{if(s.top()=='[')s.pop();else{ans++;s.pop();}}}elseflag=1;}if(l!=k){printf("Impossible\n");}else{if(flag==1)printf("Impossible\n");elseprintf("%d\n",ans);}}return 0;}
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