Brackets - POJ 1955 dp

探讨了如何通过动态规划算法解决寻找给定括号序列中最长合法子序列的问题,并提供了一个具体的AC代码示例。

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Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 3563 Accepted: 1842

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 … 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 i1i2, …, imwhere 1 ≤ i1 < i2 < … < im ≤ nai1ai2 … 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.

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6

题意:求给定序列的合法子序列的最长长度。

思路:简单的dp,直接看代码的转移吧。

AC代码如下:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[110];
int n,dp[110][110];
int main()
{
    int i,j,k,d;
    while(~scanf("%s",s+1) && s[1]!='e')
    {
        n=strlen(s+1);
        memset(dp,0,sizeof(dp));
        for(d=1;d<n;d++)
           for(i=1;i+d<=n;i++)
           {
               j=i+d;
               for(k=i;k<j;k++)
                  dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
               if((s[i]=='(' && s[j]==')')||(s[i]=='[' && s[j]==']'))
                  dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
           }
        printf("%d\n",dp[1][n]);
    }
}



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