poj 2955 Brackets(区间dp)-白红宇

poj 2955 Brackets(区间dp)

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发布时间：2019-06-22

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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_m where 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

Source

题意：选出最多的括号匹配数

区间dp

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 #include
           
             6 #include
            
              7 #include
             
               8 #include
              
                9 #include
               10 using namespace std;11 #define N 10612 #define inf 1<<2613 int n;14 char s[N];15 int dp[N][N];16 bool check(int x,int y)17 {18 if(s[x]=='(' && s[y]==')')19 return true;20 if(s[x]=='[' && s[y]==']')21 return true;22 return false;23 }24 int main()25 {26 while(scanf("%s",s) && s[0]!='e')27 {28 n=strlen(s);29 //for(int i=1;i<=n;i++)30 // printf("---%c",s[i]);31 32 memset(dp,0,sizeof(dp));33 for(int i=0;i

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