POJ-2955-Brackets

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6947 Accepted: 3727

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, and
• if 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 a₁a₂ … a_n, 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, …, i_mwhere 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

66406

dp[i][j]为串中第 i 个到第 j 个括号的最大匹配数目
那么如果第i个和第j个是一对匹配的括号那么dp[i][j] = dp[i+1][j-1]+2;
dp[i][j] = max(dp[i][j], dp[i][k]+dp[k+1][j]);

dp[][]的MAX开到120就够了，太大会T掉 ，QAQ 

[cpp] view plain copy

1. #include <iostream>  
2. #include <stdio.h>  
3. #include <string.h>  
4. #include <algorithm>  
5. #include <queue>  
6. #include <vector>  
7. #include <map>  
8. #include <cmath>  
9. #include <stdlib.h>  
10. using namespace std;  
11. const double PI = acos(-1.0);  
12. const double eps = 0.1;  
13. const int MAX = 120;  
14. const int mod = 1e9+7;  
15. int dp[MAX][MAX];  
16. string s;  
17. int main()  
18. {  
19.     while(cin>>s)  
20.     {  
21.         if(s=="end")break;  
22.         memset(dp, 0, sizeof(dp));  
23.         for(int l = 1; l<(int)s.size(); ++l)  
24.         {  
25.             for(int i = 0; i+l<(int)s.size(); ++i)  
26.             {  
27.                 int j = i+l;  
28.                 if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))  
29.                     dp[i][j] = dp[i+1][j-1]+2;  
30.                 for(int k = i; k<j; ++k)  
31.                     dp[i][j] = max(dp[i][j], dp[i][k]+dp[k+1][j]);  
32.             }  
33.         }  
34.         printf("%d\n", dp[0][(int)s.size()-1]);  
35.     }  
36.     return 0;  
37. }