Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8085		Accepted: 4299

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

 

 

 

 

根据规则来进行区间DP

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define lli long long int 
using namespace std;
const int MAXN=1001;
const int maxn=0x7fffff;
void read(int &n)
{
    char c='+';int x=0;bool flag=0;
    while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
    while(c>='0'&&c<='9')
    x=(x<<1)+(x<<3)+c-48,c=getchar();
    flag==1?n=-x:n=x;
}
char s[MAXN];
int dp[MAXN][MAXN];
int  main()
{
    while(scanf("%s",s))
    {
        if(s[0]=='e')
            break;
        memset(dp,0,sizeof(dp));
        int l=strlen(s);
        for(int i=l;i>=0;i--)
            for(int j=i;j<l;j++)
            {
                //dp[i][j]=max(dp[i+1][j],dp[i][j-1]);
                
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                    dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
                    
                for(int k=i;k<j;k++)    
                    dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]);
            
            }
        printf("%d\n",dp[0][l-1]);
    }
    return 0;
}