poj 2955 Brackets(区间dp)

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7441 Accepted: 3960

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

66406

tips:

dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j])  i<=k<=j

#include<iostream>#include<cstring>#include<string>using namespace std;string s;int dp[110][110];int judge(char c1,char c2){return ((c1=='('&&c2==')')||(c1=='['&&c2==']'));}int main(){while(cin>>s,s!="end"){memset(dp,0,sizeof(dp)); for(int r=2;r<=s.size();r++)//range{for(int i=0;i<=s.size()-r;i++)//起点 {int j=i+r-1;if(judge(s[i],s[j]))dp[i][j]=dp[i+1][j-1]+2;for(int k=i;k<=j;k++)//分割位置 {dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);} }}cout<<dp[0][s.size()-1]<<endl;}return 0; }