设为首页 加入收藏

TOP

POJ 2955 Brackets
2015-07-24 05:32:09 来源: 作者: 【 】 浏览:6
Tags:POJ 2955 Brackets

Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: Accepted:

Description

We give the following inductive definition of a “regular brackets” sequence:

the empty sequence is a regular brackets sequence,if s is a regular brackets sequence, then ( s) and [ s] are regular brackets sequences, andif a and b are regular brackets sequences, then ab is a regular brackets sequence.no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a1a2an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤i1 < i2 < … < imn, ai1ai2 … aim is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6

代码中详解!!!


AC代码如下:

#include
  
   
#include
   
     #include
    
      using namespace std; int dp[105][105]; char str[1005]; int main() { int i,j,o,t; while(cin>>str,str[0]!='e') { int l=strlen (str); for(i=0;i
     
      



】【打印繁体】【投稿】【收藏】 【推荐】【举报】【评论】 【关闭】 【返回顶部
分享到: 
上一篇HDU1392:Surround the Trees(凸.. 下一篇QT 监听 USB 设备 插入、拔出动作

评论

帐  号: 密码: (新用户注册)
验 证 码:
表  情:
内  容: