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 i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … 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
分析:
还是按照之前的方法,把子区间分割,然后如果匹配,则dp[i][j]=dp[i+1][j-1]+2’
AC code:
#include<iostream>
#include<cstring>
using namespace std;
const int maxn=105;
int dp[maxn][maxn];
int main()
{
string s;
while(cin>>s)
{
if(s=="end") break;
memset(dp,0,sizeof(dp));
for(int i=s.size()-1;i>=0;i--)
{
for(int j=i;j<s.size();j++)
{
if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[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]);
}
}
}
cout<<dp[0][s.size()-1]<<endl;
}
}