区间DP：Brackets

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本文介绍了一种使用动态规划算法解决寻找字符串中最长的有效括号序列的问题，通过枚举和状态转移方程来求解最大匹配数目。

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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 … 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<cstdio>  
#include<cstring>  
#include<cmath>  
#include<set>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
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; i<len; i++)  
            for(j=0,k=i; k<len; j++,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<k; 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<<dp[0][len-1]<<endl;  
    }  
    return 0;  
}