<> Title Description

Please write a program , Determine whether the brackets in the given expression match , The legal bracket in the expression is ”(“, “)”, “[", "]“, “{“,
”}”, The three brackets can be nested in any order .

<> input

There are multiple expressions , The first line of input data is the number of expressions , Each expression occupies one line .

<> output

For each expression , If the brackets match , Then output ”yes”, Otherwise output ”no”.

<> sample input
4 [(d+f)*{}] [(2+3)) ()} [4(6]7)9
<> sample output
yes no no no
thinking : use stack as well as map, adopt map take ”(“, “)”, “[", "]“, “{“,
”}” Any two of them are tied together . After entering a string , If the elements of the string are parentheses , Put it on the stack , If the bracket is a pair with the top element of the stack , The top of the stack element is removed from the stack . last , If the stack is empty , Is legal , If not empty , It's not legal .
#include <iostream> #include <string> #include <stack> #include <cstdio> #
include <map> using namespace std; int main() { int n,i,j; string str; stack<
char> s; map<char,char> mp; // Tie the brackets together mp['(']=')'; mp['{']='}'; mp['[']=']'; mp
[')']='('; mp['}']='{'; mp[']']='['; while(scanf("%d\n",&n)!=EOF){ for(i=0;i<n;i
++){ getline(cin,str); for(j=0;j<str.length();j++){ if(!s.empty()&&mp[s.top()]==
str[j]){ // If the stack is not empty and the character is a good friend of the top element of the stack s.pop(); // The top of the stack element is removed from the stack continue; // And skip the next step }
// If it's a bracket , Then put it on the stack if(str[j]=='('||str[j]=='{'||str[j]=='['||str[j]==')'||str[j]=='}'||
str[j]==']') s.push(str[j]); } if(s.empty()) printf("yes\n"); // Empty stack means legal else
printf("no\n"); // If the stack is not empty, it is illegal while(!s.empty()) s.pop(); // At this time, if the stack is not empty , There will be residues in the stack , The stack should be empty }
} return 0; }

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