poj 2955 区间dp

题目链接：https://vjudge.net/problem/23652/origin

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

6
6
4
0
6

输入一个字符串，我们存进str数组，问你最多有多少个合法括号（"()"这里是两个），在这里我们可以先把字符串的长度算出来为len，那么问题就是在
0到len-1的区间里
最多有多少个合法
括号，我们定义一个dp数组，dp[i][j]表示在i到j的区间里最多有多少个合法括号，step1：在str[i]==str[j]时，dp[i][j]=dp[i+1][j-1]+2;   
然后 step2：我们就在[i,j)(左闭右开区间)里把每一个数设成k，用k来分割区间，把它分割为两个更小的区间，如果dp[i][j]<dp[i][k]+dp[k+1][j]
那么就更新dp[i][j]的值，
注意：step1和step2是都要算的，解释来自：博客链接：https://www.cnblogs.com/Silenceneo-xw/p/5939751.html，因为如果不算step2
那么就会出现"[][]"转移到"]["的情况，如果不算step1，那么答案为0，其实在上面step2里我们不是一个[i,j)的开区间吗？其实我个人认为可以把
step1看成上面开区间的补充(把k=j的情况单独列出,只是这个j有点特别，要特别对待)，就是j也看成k，这样k所在区间就是一个左闭右闭区间[i,j]了。
看代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char str[105];
int dp[105][105],n,m,k,t;
int main()
{
    while(fgets(str,105,stdin))
    {
        if(str[0]=='e')
        break;
        int len=strlen(str);
        memset(dp,0,sizeof(dp));
        for(int l=1;l<len;l++)
        {
            for(int i=0;i<len-l;i++)
            {
                int j=i+l;
                if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']'))
                dp[i][j]=dp[i+1][j-1]+2;
                for(int k=i;k<j;k++)
                {
                    if(dp[i][j]<dp[i][k]+dp[k+1][j])
                    dp[i][j]=dp[i][k]+dp[k+1][j];
                }
            }
        }
        printf("%d
",dp[0][len-1]);
    }
    return 0;
 }