栈的应用（平衡括号）

Description

There are six kinds of brackets: ‘(‘, ‘)’, ‘[‘, ‘]’, ‘{’, ‘}’. dccmx’s girl friend is now learning java programminglanguage, and got mad with brackets! Now give you a string of brackets. Is it valid? For example: “(([{}]))” is valid, but“([)]” is not.

Input

 First line contains an integer T (T<=10): the number of test case.

Next T lines, each contains a string: the input expression consists of brackets.

The length of a string is between 1 and 100.

Output

For each test case, output “Valid” in one line if the expression is valid, or “Invalid” if not.

Sample Input

 

2
{{[[(())]]}}
({[}])

Sample Output 

Valid
Invalid

代码：

  #include<stdio.h>
        #include<string.h>
        int main()
      {
      char stack(char a[],int n);
      int n,i,len,j;
      char a[100];
     scanf("%d%*c",&n);
     for(i=0;i<n;i++)
     {
       scanf("%s",a);
       len=strlen(a);
       if(stack(a,len))
       printf("Valid\n");
       else
        printf("Invalid\n");
     }
     return 0;
     }
     char stack(char a[],int n)
     {
      int i;
      char b[100],t;
      int top=0,buttom=0;
      for(i=0;i<n;i++)
     {
         if(a[i]=='(' || a[i]=='['  || a[i]=='{')
         b[top++]=a[i];   
       else
      {
       if(top==0)
       {top=1;
        break;}
         t=b[top-1];
         if((t=='['&&a[i]==']')||(t=='{'&&a[i]=='}')||(t=='('&&a[i]==')'))
        {    --top;   t=b[top-1];   }
      }
     }
        if(top==buttom)
        return 1;
       else return 0; 

      }