Brackets - POJ 1955 dp

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal 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 … 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_mwhere 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

66406

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]);    }}