poj 2955 Brackets(区间dp)

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

 

题意：选出最多的括号匹配数

 

      区间dp

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<algorithm>
#include<cmath>
#include<stdlib.h>
#include<map>
using namespace std;
#define N 106
#define inf 1<<26
int n;
char s[N];
int dp[N][N];
bool check(int x,int y)
{
    if(s[x]=='(' && s[y]==')')
      return true;
     if(s[x]=='[' && s[y]==']')
       return true;
     return false;
}
int main()
{
    while(scanf("%s",s) && s[0]!='e')
    {
        n=strlen(s);
        //for(int i=1;i<=n;i++)
          // printf("---%c",s[i]);
        
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
          {
              if(check(i,i+1))
                dp[i][i+1]=2;
          }
        for(int len=1;len<n;len++)
        {
            for(int i=0;i+len<n;i++)
            {
                int j=i+len;
                if(check(i,j))
                  dp[i][j]=dp[i+1][j-1]+2;
                  for(int k=i;k<=j;k++)
                  {
                      dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                  }
            }
        }
        printf("%d
",dp[0][n-1]);
        
    }
    return 0;
}