POJ2955 Brackets （区间DP）

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

6
6
4
0
6
区间DP经典例题。因为括号是可以嵌套并列的，所以想到区间DP（才不是因为看到了标签..这个DP数组的含义比较好想到,dp[i][j]表示下标i~j的这一段字符串最多有多少个括号匹配了，按照区间DP的常见套路，第一层枚举区间长度，由于匹配最少需要两个括号，所以len从2开始；第二层枚举左端点，同时也可以根据左端点和长度确定右端点。这里先判断一步，如果s[i]和s[j]匹配了，那么dp[i][j]起码是dp[i+1][j-1]+2（为什么说是起码呢？因为考虑样例的()()()这个序列自然就明白了）
然后找决策点，看看由两个分开的子区间能不能转移到当前并且使得答案更大。最后输出的就是dp[0][len-1]。

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
char s[105];
int dp[105][105];
bool check(int l,int r)
{
    if(s[l]=='('&&s[r]==')')return true;
    if(s[l]=='['&&s[r]==']')return true;
    return false;
}
int main()
{
    while(scanf("%s",s)&&s[0]!='e')
    {
        memset(dp,0,sizeof(dp));
        int len,i,j,k;
        for(len=2;len<=strlen(s);len++)
        {
            for(i=0;i+len-1<=strlen(s)-1;i++)
            {
                int j=i+len-1;
                if(check(i,j))
                {
                    dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
                }
                for(k=i;k<=j-1;k++)
                {
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
                
            }
        }
        cout<<dp[0][strlen(s)-1]<<endl;
    }
    return 0;
}