Uva 673平衡的括号
作者:互联网
Uva 673平衡的括号
题目描述:
思路:
就是普通的括号匹配问题,用栈来模拟操作。需要注意的地方是,有输入为空的情况,所以不要用cin
来读取,而是用getline
。
代码:
#include <iostream>
#include <stack>
#include <string>
#include <sstream>
using namespace std;
int main()
{
//freopen("uva673_in.txt", "r", stdin);
//freopen("uva673_out.txt", "w", stdout);
string s;
getline(cin, s);
stringstream ss(s);
int n ; ss >> n;
while(n--){
string s; getline(cin, s);
stack<char> res;
int failed = 0;
for(int i = 0; i < s.length(); ++i){
if(s[i] == ')'){
if(!res.empty()) {
char c = res.top();
if(c == '(') res.pop();
else { failed = 1; break;}
}
else { failed = 1; break;}
}else if(s[i] == ']'){
if(!res.empty()){
char c = res.top();
if(c == '[') res.pop();
else { failed = 1; break;}
}
else { failed = 1; break;}
}
else res.push(s[i]);
}
if(res.empty() && !failed) cout << "Yes\n";
else cout << "No\n";
}
}
标签:include,failed,int,res,else,break,括号,Uva,673 来源: https://www.cnblogs.com/patrolli/p/11291083.html