POJ 2955：Brackets（区间DP）_brackets poj

本文链接：https://blog.youkuaiyun.com/xiaochengxvyuan/article/details/125548845

该博客讨论了一种动态规划解决方案，用于寻找给定字符串中最大长度的有效括号子序列。输入是一个包含括号的字符串，任务是找到最长的可以组成合法括号配对的子串。区间动态规划策略被用来遍历字符串并计算不同子串的有效性，最终输出最长有效括号子序列的长度。示例代码展示了如何实现这一算法。

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题目

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	Output
((())) ()()() ([]]) )[)( ([][][) end	6 6 4 0 6

题目大意：给你一个字符串，找最大能凑成对的括号序列

思路：区间dp

状态表示 f[i,j] 表示从i到j的区间里合法的成对的括号序列长度

属性：最大值

状态计算：

1 . 当第i个和第j个匹对时，让 f[i,j] = f[i+1,j-1] +2；

如果不匹对 f[i,j] = f[i+1,j-1]

2 ，但是不能保证上方操作就是最大值，在i，j 区间里，找怎么合并取最大最大值，

代码如下

#include<iostream>
#include<cstring>
using namespace std;
const int N=111;
int f[N][N];
struct node  // 记录当前的符号 
{
	bool n1,n2,m1,m2;
}a[N];
string s;

int main()
{
	while(cin>>s)
	{
		memset(a,0,sizeof a);
		if(s=="end") break;
		
		int n=s.size();
		for(int i=0;i<n;i++)
		{
			if(s[i]=='[') a[i+1].m1=true;
			if(s[i]==']') a[i+1].m2=true;
			if(s[i]=='(') a[i+1].n1=true;
			if(s[i]==')') a[i+1].n2=true;
		}
		memset(f,0,sizeof f);    //  初始化 
		for(int len=2;len<=n;len++)
		{
			for(int l=1;l<=n-len+1;l++)
			{
				int r=len+l-1;
				if((a[l].m1 && a[r].m2 )||( a[l].n1 && a[r].n2))  //  如果匹对成功 
				{
					f[l][r]=f[l+1][r-1]+2;
				}
				else f[l][r]=f[l+1][r-1];
				
				for(int k=l;k<r;k++)  //  在区间中找最大值 
				{
					f[l][r]=max(f[l][r],f[l][k]+f[k+1][r]);
				}
			}
		}
		cout<<f[1][n]<<endl;
	}
	
	
	return 0;
}