UVA - 673 - Parentheses Balance （棧的應用！）

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UVA - 673

Parentheses Balance

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Parentheses Balance

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

(a)

if it is the empty string

(b)

if A and B are correct, AB is correct,

(c)

if A is correct, (A ) and [A ] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input The file contains a positive integer  n and a sequence of  n strings of parentheses () and [], one string a line.

Output A sequence of Yes or No on the output file.

Sample Input 

3([])(([()])))([()[]()])()

Sample Output 

YesNoYes

Miguel Revilla
2000-08-14

Source

Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 2. Data Structures and Libraries :: Data Structures With Built-in Libraries :: STL stack
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Data Structures and Libraries :: Linear Data Structures with Built-in Libraries :: C++ STL stack (Java Stack)
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Rare Topics :: Rare Problems :: Bracket Matching
Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 2. Data Structures :: Lists
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 6. Data Structures :: Exercises

簡單棧的應用。。判斷括弧是否匹配

沒看到Null 字元，WA了兩次。。

AC代碼：

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <stack>using namespace std;int main(){int n;while(scanf("%d", &n) != EOF){getchar();while(n--){char a[130];stack<char> s;gets(a);int flag = 0;for(int i=0; a[i] != '\0'; i++){if(a[i] == '(' || a[i] == '[') s.push(a[i]);else if(a[i] == ')') {if(!s.empty() && s.top() == '(') s.pop();else { flag = 1; break; }}else if(a[i] == ']'){if(!s.empty() && s.top() == '[') s.pop();else { flag = 1; break; }}}if(flag || !s.empty()) printf("No\n");else printf("Yes\n");}}return 0;}

UVA - 673 - Parentheses Balance （棧的應用！）