目录

Brackets (区间DP)

题目

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 … an, 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 < … < im ≤ n, 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

题意

输出能够匹配的最多的括号个数

思路

由小区间的数量推出大区间的括号数量

1. 对于小区间 如果能够首尾匹配那么 dp[i][j]=dp[i+1][j-1]+2 因为这是最直接的。
2. 但是这样不是最准确的。
3. 对于 ()() 来说 dp[2][3]=0 因为 )( 不匹配。那么用一算出来的 dp[1][4]=0+2=2 是错的，其应该等于 4 .
4. 所以这就应该进行扫描子区间。 实际最大值应该是
   dp[1][4] = max( dp[1][4] , dp[1][2] + dp[3][4] ) = 4

题解

#include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;int d[105][105];int main(){    char s[105];    while (scanf("%s", s + 1), s[1] != 'e')    {        memset(d, 0, sizeof d);        int len = strlen(s + 1);        //先枚举小区间，因为后面计算大区间时需要使用小区间        for (int l = 0; l <= len; l++)        {            for (int i = 1; i + l - 1 <= len; i++)            {                int j = l + i - 1;                //基本的状态转移                if ((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))                {                    d[i][j] = d[i + 1][j - 1] + 2;                }                //为了找到真正最大的值，进行扫描。                for (int k = i; k < j; k++)                 {                    d[i][j] = max(d[i][j], d[i][k] + d[k + 1][j]);                }            }        }        cout << d[1][len] << endl;    }}