CDZSC_2015寒假新人（4）——搜索

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 

Yes
No
Yes

---

Miguel Revilla
2000-08-14

 

题解：栈。括号匹配问题，左括号全压栈，右括号若是与栈顶匹配，则出栈顶，否则压栈，注意输入有可能是空串，要用 gets() 或 cin.getline() 输入，这坑了我一下，英语文盲的悲剧。详见代码。

 

#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;

const int MAX = 256;
char str[MAX];
int Map[MAX];


int main()
{
#ifdef CDZSC_OFFLINE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    Map['('] = -1; Map[')'] = 1;
    Map['['] = -2; Map[']'] = 2;

    int t, len;
    scanf("%d", &t);
    getchar();
    while(t--)
    {
        gets(str);

        len = strlen(str);
        stack<char> st;
        for(int i = 0; i < len; i++)
        {
            if(str[i] == '(' || str[i] == '[')
            {
                st.push(str[i]);
            }
            else
            {
                if(!st.empty() && !(Map[st.top()] + Map[str[i]]))
                {
                    st.pop();
                }
                else
                {
                    st.push(str[i]);
                }
            }
        }
        printf("%s
", st.empty() ? "Yes" : "No");
    }
    return 0;
}