CDZSC_2015寒假新人（4）——搜索 - P

Description

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

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。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stack>
 4 using namespace std;
 5
 6 const int MAX = 256;
 7 char str[MAX];
 8 int Map[MAX];
 9
10
11 int main()
12 {
13 #ifdef CDZSC_OFFLINE
14     freopen("in.txt", "r", stdin);
15     freopen("out.txt", "w", stdout);
16 #endif
17     Map[‘(‘] = -1; Map[‘)‘] = 1;
18     Map[‘[‘] = -2; Map[‘]‘] = 2;
19
20     int t, len;
21     scanf("%d", &t);
22     getchar();
23     while(t--)
24     {
25         gets(str);
26
27         len = strlen(str);
28         stack<char> st;
29         for(int i = 0; i < len; i++)
30         {
31             if(str[i] == ‘(‘ || str[i] == ‘[‘)
32             {
33                 st.push(str[i]);
34             }
35             else
36             {
37                 if(!st.empty() && !(Map[st.top()] + Map[str[i]]))
38                 {
39                     st.pop();
40                 }
41                 else
42                 {
43                     st.push(str[i]);
44                 }
45             }
46         }
47         printf("%s\n", st.empty() ? "Yes" : "No");
48     }
49     return 0;
50 }