UVA673 Parentheses Balance

题目链接： http://acm.hust.edu.cn/vjudge/contest/view.action?cid=43041#problem/E

题目大意：给你一个字符串，让你判断是否能将里面的括号按照顺序配对，例如(){},或者(()[]),但像([)]这样就是不可以的。

解题思路：建立一个栈，碰到左括号就进栈，右括号则出栈并匹配。

题目坑点：

1.注意到空字符串了没？

2.出栈之前判断是否为空栈了没？

3.字符串读完以后判断栈空了没？

代码如下：

#include <iostream>#include <stdio.h>#include <fstream>#include <iomanip>#include <cmath>#include <string>#include <string.h>#include <sstream>#include <cctype>#include <climits>#include <set>#include <map>#include <queue>#include <vector>#include <iterator>#include <algorithm>#include <stack>#include <functional>/*int类型最大值INT_MAX，short最大值为SHORT_MAXlong long最大值为LONG_LONG_MAX*///cout << "OK" << endl;#define _clr(x) memset(x,0,sizeof(x))using namespace std;const int MAXN = 200;const int INF = INT_MAX;const double eps = 1e-8;const double EULER = 0.577215664901532860;const double PI = 3.1415926535897932384626;const double E = 2.71828182845904523536028;typedef long long LL;int main(){    //freopen("sample.in", "r", stdin);//freopen("sample.out", "w", stdout);int t;cin >> t;cin.ignore(100,'\n');while(t--){stack<char>x;string s;int p = 1;getline(cin,s);if(s == "") {cout << "Yes" << endl;continue;}//cout << strlen(s) << endl;for(int i = 0;i<s.length();i++){if(s[i] == '(' || s[i] == '[')x.push(s[i]);else{if(x.empty()){p = 0;break;}char ch = x.top();x.pop();if( (s[i] == ')' && ch != '(') || (s[i] == ']' && ch != '[') ){p = 0;break;}}}if(!x.empty())p = 0;if(p)cout << "Yes" << endl;elsecout << "No" << endl;}    //fclose(stdin);    //fclose(stdout);     return 0;}