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最新推荐文章于 2020-02-07 11:05:28 发布

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本文介绍了一种通过动态规划算法来解决寻找字符串中最长的有效括号子序列的问题，并提供了详细的实现代码。该算法通过判断每两个相邻字符是否能形成有效的括号对，进而更新状态数组以找出最长的有效括号子序列长度。

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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

先判断每两个相邻的字符是否能配成括号，然后枚举各个区间的情况

如果s[i]和s[j]能配成括号则dp[i][j]=dp[i+1][j-1]+2

更新区间[i][j]的最大值 dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]);

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    char s[110];
    int n,dp[110][110];

    while(~scanf("%s",s))
    {
        if(strcmp(s,"end")==0) break;
        n=strlen(s);
        memset(dp,0,sizeof(dp));

        for(int j=1; j<n; j++)
        {
            for(int i=j-1; i>=0; i--)
            {
                if((s[i]=='(' && s[j]==')') || (s[i]=='[' && s[j]==']'))
                    dp[i][j]=dp[i+1][j-1]+2;
                for(int k=i; k<=j; k++)
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
            }
        }
        printf("%d\n",dp[0][n-1]);
    }
    return 0;
}