程序设计思维与实践 Week12 作业 D 选做题 - 1

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.

题目简述：

给定一串()[]的字符串，从中选取最长的能够实现“正则”（括号合法匹配）的序列，输出序列长度。多组数据，end结束。

思路：

区间dp。f[i][j]表示从i到j符合题意的子序列的最大长度。因为枚举时，较长的区间需要从较短的区间转移，因此首先枚举区间长度。然后从头开始枚举起点i，因为长度已知了，所以j可以确定。如果i j位置处恰好可以匹配：s[i]=='(' s[j]==')'或者s[i]=='[' s[j]==']'，所以此时f[i][j]=f[i+1][j-1]+2.现在考虑区间内部，ij序列可以由包含在ij序列内的子序列i~k k+1~j得到，因此需要再次更新。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
string s;
int n,f[1010][1010];
bool judge(int i,int j)
{
	if(s[i]=='['&&s[j]==']') return 1;
	if(s[i]=='('&&s[j]==')') return 1;
	return 0;
}
int main()
{
	while(1)
	{
		cin>>s;
		if(s=="end") break;
		n=s.length();
		memset(f,0,sizeof(f));
		for(int l=1;l<n;l++)
		for(int i=0;i+l<n;i++)
		{
			int j=i+l;
			if(judge(i,j))
			{
				if(j==i+1) f[i][j]=2;
				else f[i][j]=f[i+1][j-1]+2;
			}
			for(int k=i;k<j;k++)
			f[i][j]=max(f[i][j],f[i][k]+f[k+1][j]);
		}
		cout<<f[0][n-1]<<endl;
	}
	return 0;
}