poj2955 Brackets (区间dp)

最新推荐文章于 2019-03-14 09:30:44 发布

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最新推荐文章于 2019-03-14 09:30:44 发布

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分类专栏： DP 动态规划

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Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3571		Accepted: 1847

Description

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

Source

Stanford Local 2004

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 105

char a[N];
int dp[N][N];

int main()
{
	while(scanf("%s",a)&&a[0]!='e')
	{
		 memset(dp,0,sizeof(dp));
		 int len=strlen(a);
		 for(int i=len-1;i>=0;i--)
			for(int j=i+1;j<len;j++)
		 {
		 	dp[i][j]=dp[i+1][j];
		 	for(int k=i+1;k<=j;k++)
				if(a[i]=='('&&a[k]==')'||a[i]=='['&&a[k]==']')
				  dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k][j]+2);
		 }
		 printf("%d\n",dp[0][len-1]);
	}
	return 0;
}