括号匹配，栈的基本应用；

样题：

Brackets! Brackets!

Time Limit: 1000 MS Memory Limit: 65536 K Total Submit: 811(283 users) Total Accepted: 333(256 users) Rating: Special Judge: No

Description

There are six kinds of brackets: ‘(‘, ‘)’, ‘[‘, ‘]’, ‘{’, ‘}’. dccmx’s girl friend is now learning java programming language, 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 这是典型的栈的基础应用--括号匹配； 详细思路在代码注释里； #include #include #include #include #include using namespace std; int main() {     char c[1000];     int n,i,d = 1;     scanf("%d",&n);     while(n--)     {         scanf("%s",&c);         int str = strlen(c);         stack st;         for(i = 0; i < str; i++)         {             if(c[i] == '(' || c[i] == '{'||c[i] == '[')                 st.push(c[i]);             else if(c[i]==')'&&!st.empty()&&st.top()=='(')///遇到右括号时，当栈不为空时，栈首一定是'(';                 st.pop();///左括号找到匹配的，可以弹出；             else if(c[i]==')'&&!st.empty()&&st.top()!='(')///这是不符合规则的，做一个标记；                 d=0;             else if(c[i]==')'&&st.empty())///这是判断有没有没出现左括号，就有右括号的情况；                 d=0;///因为入栈的只能是左括号；             else if(c[i] == ']'&&!st.empty()&&st.top()=='[')                 st.pop();             else if(c[i] == ']'&&!st.empty()&&st.top()!='[')                 d=0;             else if(c[i]==']'&&st.empty())                 d=0;             else if(c[i]=='}'&&!st.empty()&&st.top()=='{')                 st.pop();             else if(c[i]=='}'&&!st.empty()&&st.top()!='{')                 d=0;             else if(c[i]=='}'&&st.empty())                 d=0;         }         if(!st.empty()||d == 0)///必须要弹尽栈里的元素，且不能有不符合规则的；             printf("Invalid\n");         else             printf("Valid\n");         d = 1;     }     return 0; }