链接:http://acm.hust.edu.cn/vjudge/problem/19102分析:特别要注意空串情况,还有括号奇数个肯定不合法都可以特判。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <stack> 5 using namespace std; 6 7 int main() { 8 int n; 9 cin >> n; getchar(); 10 while (n--) { 11 string line; 12 getline(cin, line); 13 if (line.compare("\n") == 0) { 14 printf("Yes"); continue; 15 } 16 if (line.size() % 2) { 17 printf("No\n"); continue; 18 } 19 stack<char> s; 20 bool ok = true; 21 for (int i = 0; i < line.size(); i++) { 22 char ch = line[i]; 23 if (ch == ‘(‘) s.push(ch); 24 else if (ch == ‘[‘) s.push(ch); 25 if (ch == ‘)‘) 26 if (!s.empty() && s.top() == ‘(‘) s.pop(); 27 else { ok = false; break; } 28 else if (ch == ‘]‘) 29 if (!s.empty() && s.top() == ‘[‘) s.pop(); 30 else { ok = false; break; } 31 } 32 if (ok && s.empty()) printf("Yes\n"); 33 else printf("No\n"); 34 } 35 return 0; 36 }
时间: 2024-11-10 08:25:06