题干：

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

解题报告：

    水题模拟不解释。就是找到关键元素在于，对于第一个出现的右括号，与之相邻的一定是左括号，抓住这一点，程序很好写了。

AC代码：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
bool iszuo(char ch) {return ch=='(' || ch=='[' || ch=='{';
}
bool isyou(char ch) {return ch==')' || ch==']' || ch=='}';
} 
char s[1005];
int main() 
{int t;cin>>t;getchar();while(t--) {stack<char> sk;//getline(cin,s);cin.getline(s,1005);int len = strlen(s);int flag = 1;for(int i = 0; i<len; i++) {if(iszuo(s[i])) sk.push(s[i]);else {if(sk.empty()) {flag=0;break;}if(s[i] ==')') {if(sk.top() != '(') {flag=0;break;}}else if(s[i] == ']') {if(sk.top() != '[') {flag=0;break;}}else {if(sk.top() != '{') {flag=0;break;}}sk.pop();}}if(flag && sk.empty()) puts("Valid");else puts("Invalid");}return 0;
}

总结：

   虽然是一道水题但是还是有很多值得总结的地方。比如要时刻看看是否栈为空，包括最后输出的地方，因为栈不空时，显然是invalid的情况！

   还有就是因为整个就在用sk.top()和s[i]这两个值，所以别用错了，比如有个地方的sk.top()我就写成了s[i]