区间DP：Brackets

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 … 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

定义dp [ i ] [ j ] 为串中第 i 个到第 j 个括号的最大匹配数目
那么我们假如知道了 i 到 j 区间的最大匹配，那么i+1到 j+1区间的是不是就可以很简单的得到。
那么 假如第 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 ;
那么我们只需要从小到大枚举所有 i 和 j 中间的括号数目，然后满足匹配就用上面式子dp，然后每次更新dp [ i ] [ j ]为最大值即可。
更新最大值的方法是枚举 i 和 j 的中间值，然后让 dp[ i ] [ j ] = max ( dp [ i ] [ j ] , dp [ i ] [ f ] + dp [ f+1 ] [ j ] ) 。

#include  
#include  
#include  
#include  
#include  
#include  
#include  
using namespace std;  
#define MAXN 202  
#define INF 999999  
int dp[MAXN][MAXN];  

int main()  
{  
    string s;  
    while(cin>>s)  
    {  
        if(s=="end")  break;  
        memset(dp,0,sizeof(dp));  
        int i,j,k,f,len=s.size();  
        for(i=1; ifor(j=0,k=i; k//枚举区间长度，区间端点
            {  
                if((s[j]=='('&&s[k]==')')||(s[j]=='['&&s[k]==']'))  
                    dp[j][k]=dp[j+1][k-1]+2;  
                for(f=j; f//划分更新区间
                    dp[j][k]=max(dp[j][k],dp[j][f]+dp[f+1][k]);  
            }  
        //len-dp[0][len-1]表示需要补充多少括号才能都匹配成功  
        cout<0][len-1]<return 0;  
}