poj 2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3961		Accepted: 2085

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

6
6
4
0
6

Source

Stanford Local 2004

题目链接： http://poj.org/problem?id=2955
【题意】最多多少个匹配的
【思路】区间dp，dp[i][j]表示第i个到第j个最多多少个匹配.
                                   dp[i][j]=min(dp[i][j],dp[i+1][j-1]+1)  s[i]==s[j];
                                   dp[i][j]=min(dp[i][j],dp[i]][k]+dp[k+1][j]) i 【代码】

/*************************************************************************
    > File Name: poj2955.cpp
    > Author: wanghao
    > Mail: [email protected] 
    > Created Time: 2015年07月09日 星期四 22时52分19秒
 ************************************************************************/

#include
#include
#include
#define ll long long
using namespace std;

int dp[110][110];
int judge(char a,char b)
{
	if(a=='('&&b==')')
		return 1;
	if(a=='['&&b==']')
		return 1;
	return 0;
}
int main()
{
	char s[110];
	while(scanf("%s",s+1)!=EOF)
	{
		if(strcmp(s+1,"end")==0)
			break;
		int len=strlen(s+1);
		memset(dp,0,sizeof(dp));
		for(int l=1;l<=len;l++)
		{
			for(int i=1;i+l-1<=len;i++)
			{
				int j=i+l-1;
				if(judge(s[i],s[j]))
					dp[i][j]=dp[i+1][j-1]+1;
				for(int k=i;k<=j;k++)
					dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]);

			}
		}
		cout<<2*dp[1][len]<<'\n';
	}
	return 0;
}