Brackets_区间DP

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6217		Accepted: 3331

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

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 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<iostream>
#include<stdio.h>
#include<string>
#include<string.h>
using namespace std;
const int N=120;
int dp[N][N];
int main()
{
    string s;
    while(cin>>s)
    {
        if(s=="end") break;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<s.size();i++)
        {

            for(int j=0,k=i;k<s.size();j++,k++)
            {
                if(s[j]=='('&&s[k]==')'||s[j]=='['&&s[k]==']')
                    dp[j][k]=dp[j+1][k-1]+2;
                for(int f=j;f<k;f++)
                    dp[j][k]=max(dp[j][k],dp[j][f]+dp[f+1][k]);
            }
        }
        cout<<dp[0][s.size()-1]<<endl;
    }
    return 0;
}