POJ - 2955（2020.4.19训练C题）_we give the following inductive definition of a “r-优快云博客

本文探讨了如何在给定的包含括号的序列中找到最长的有效括号子序列的长度，采用区间动态规划方法，通过枚举起点和终点来更新最长合法序列长度。

Problem
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.

题意：给定一个序列仅含’{‘或’}‘或’[‘或’]’,求其子串中的最长合法长度为多少

区间dp题，dp[i][j]代表下标i到j的合法最长长度，列举起点j和终点k，如果s[j]=s[k],则dp[j][k]=dp[j+1][k-1]+2,然后对段中分割，更新最大值

AC代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int dp[105][105];
char s[105];
int main()
{
	int i, j, l, k;
	while (scanf("%s",s)!=EOF)
	{
		
		if (strcmp(s, "end") == 0)
		{
			return 0;
		}
		int len = strlen(s);
		//cout << len << endl;
		memset(dp, 0, sizeof(dp));
		for (i = 1; i < len; i++)
		{
			for (j = 0; j + i < len; j++)
			{
				k = i + j;
				if ((s[j]=='['&&s[k]==']')||(s[j]=='('&&s[k]==')'))
				{
					//cout << "*" << endl;
					dp[j][k] = dp[j + 1][k - 1] + 2;
				}
				for (l = j; l < k; l++)
				{
					dp[j][k] = max(dp[j][k], dp[j][l] + dp[l+1][k]);
				}
			}

		}
		cout << dp[0][len - 1] << endl;
	}
}