UVa 673 - Parentheses Balance-优快云博客

本文详细介绍了UVa673-ParenthesesBalance问题的解决思路，采用栈数据结构实现括号匹配算法，并通过具体代码实例进行演示。包括输入输出规范及样例分析。

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UVa 673 - Parentheses Balance

1 题目
2 思路
3 代码
4 参考

1题目

===================

Parentheses Balance

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

(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

Yes
No
Yes

Miguel Revilla
2000-08-14

===================

2思路

括号匹配问题，使用栈实现。需要注意的是除了检查括号是否匹配，还要检查是否完全匹配，即最后栈是否为空，这点容易出错。

3代码

/* 
 * Problem: UVa 673 - Parentheses Balance
 * Lang: ANSI C
 * Time: 0.029s
 * Author: minix
 */
#include <stdio.h>
#include <string.h>
#define N 128+10
char str[N];
char stack[N];
int top;
int main() {
  int n, i, j, len;
  scanf ("%d", &n); getchar();
  for (i=0; i<n; i++) {
    memset (str, 0, sizeof(str[0])*N);
    top = -1;
    fgets (str, sizeof(str[0])*N, stdin);
    len = strlen(str)-1;
    for (j=0; j<len; j++) {
      if (str[j]=='(' || str[j]=='[') stack[++top] = str[j];
      else {
        if (top>=0 && str[j]==')' && stack[top]=='(') 
          top--;
        else if (top>=0 && str[j]==']' && stack[top]=='[')
          top--;
        else
          break;
      }
    }
    if (j==len && top==-1) printf ("Yes\n");
    else printf ("No\n");
  }
  return 0;
}