CodeForces 612C Replace To Make Regular Bracket Sequence
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题意:
给你一个只含有括号的字符串,你可以将一种类型的左括号改成另外一种类型,右括号改成另外一种右括号
问你最少修改多少次,才能使得这个字符串匹配,输出次数
思路:
用stack,每次将左括号压进stack里面,遇到右括号就判断一下就好了
非法就很简单,看看栈最后是否还有,看看右括号的时候,左括号的栈是否为空
#include<bits/stdc++.h>using namespace std;string s;stack<char> S;int main(){ cin>>s; int ans = 0; for(int i=0;i<s.size();i++) { if(s[i]==']') { if(S.size()==0)return puts("Impossible"); if(S.top()=='[') S.pop(); else { ans++; S.pop(); } } else if(s[i]==')') { if(S.size()==0)return puts("Impossible"); if(S.top()=='(') S.pop(); else { ans++; S.pop(); } } else if(s[i]=='>') { if(S.size()==0)return puts("Impossible"); if(S.top()=='<') S.pop(); else { ans++; S.pop(); } } else if(s[i]=='}') { if(S.size()==0)return puts("Impossible"); if(S.top()=='{') S.pop(); else { ans++; S.pop(); } } else S.push(s[i]); } if(S.size()!=0)return puts("Impossible"); cout<<ans<<endl;}
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