区间dp_2

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

66406

#include<iostream>#include<cstdio>#include<cstring>using namespace std;char a[110];int dp[110][110];//dp[i][j] 表示 区间i~j有多少可匹配的括号int main(){    while(~scanf("%s",a) && a[0]!='e')    {        memset(dp,0,sizeof(dp));        int len;        len=strlen(a);        for(int i=len-1;i>=0;i--)        {            for(int j=i+1;j<=len-1;j++)            {                for(int k=i;k<=j;k++)                {                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);//三个for循环找出从i到j的最优解                }                //但是如果a[i]与a[j]匹配 还需要增加判断来确保最优。                if(((a[i]=='(')&&(a[j]==')'))||((a[i]=='[')&&(a[j]==']')))                    dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);//在这里才增加            }        }        printf("%d\n",dp[0][len-1]);    }    return 0;}