的S最长子是平衡 [英] Longest subsequence of S that is balanced

问题描述

考虑的问题：

一个字符串，括号被认为是 平衡，如果字符串中的左，右括号可以适当地配对了。例如，字符串（（））和（）（）都是平衡的，而字符串（（）（是不 均衡。
给定一个字符串的取值长度 N 括号组成的，假设你想找到的最长的子序列的取值这是平衡的。使用动态规划，设计一个算法，发现的最长的平衡子的取值在为O（n ^ 3）的时间。

Given question:

A string of parentheses is said to be balanced if the left- and right-parentheses in the string can be paired off properly. For example, the strings "(())" and "()()" are both balanced, while the string "(()(" is not balanced.
Given a string S of length n consisting of parentheses, suppose you want to find the longest subsequence of S that is balanced. Using dynamic programming, design an algorithm that finds the longest balanced subsequence of S in O(n^3) time.

我的做法：
假设给定的字符串：S [1 2 ... N]
有效的子序列可以在S [I]当且仅当S [I] =='）'结尾，即S [i]为一个右括号，并至少​​存在一个未使用的开括号previous到S [1]。它可以在O（N）来实现。

My approach:
Suppose given string: S[1 2 ... n]
A valid sub-sequence can end at S[i] iff S[i] == ')' i.e. S[i] is a closing brace and there exists at least one unused opening brace previous to S[i]. which could be implemented in O(N).

#include<iostream>
using namespace std;
int main(){
    string s;
    cin >> s;
    int n = s.length(), o_count = 0, len = 0;
    for(int i=0; i<n; ++i){
        if(s[i] == '('){
            ++o_count;
            continue;
        }
        else if(s[i] == ')' && o_count > 0){
            ++len;
            --o_count;
        }
    }
    cout << len << endl;
    return 0;
}

我尝试了几个测试案例，他们似乎是工作的罚款。我失去了一些东西呢？如果不是，那我怎么才能还设计了为O（n ^ 3）动态规划为这个问题的解决？_爱 这是子我使用的定义。

I tried a couple of test cases and they seem to be working fine. Am I missing something here? If not, then how can I also design an O(n^3) Dynamic Programming solution for this problem?

This is the definition of subsequence that I'm using.

谢谢！

推荐答案

有关为O（n ^ 3） DP这应该工作，我认为：

For O(n^3) DP this should work I think:

dp[i, j] = longest balanced subsequence in [i .. j]
dp[i, i] = 0
dp[i, i + 1] = 2 if [i, i + 1] == "()", 0 otherwise

dp[i, j] = max{dp[i, k] + dp[k + 1, j] : j > i + 1} in general

这可以实现类似于如何最优的矩阵链乘法是。

This can be implemented similar to how optimal matrix chain multiplication is.

您算法还似乎是正确的我，看到例如这个问题：

Your algorithm also seems correct to me, see for example this problem:

<一个href="http://xorswap.com/questions/107-implement-a-function-to-balance-parentheses-in-a-string-using-the-minimum-nu" rel="nofollow">http://xorswap.com/questions/107-implement-a-function-to-balance-parentheses-in-a-string-using-the-minimum-nu

当解决方案基本都是一样的你。

Where the solutions are basically the same as yours.

您只忽略了额外的支架，所以我不明白为什么它是行不通的。

You are only ignoring the extra brackets, so I don't see why it wouldn't work.

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