Description
We give the following inductive definition of 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
#include<iostream> #include<string> using namespace std; int d[105][105]; int n; string s; void dp(){ for(int i=0;i<n;i++){d[i+1][i]=0;d[i][i]=1;} for(int i=n-2;i>=0;i--){ for(int j=i+1;j<n;j++){ d[i][j]=n; if(s[i]==‘(‘&&s[j]==‘)‘||s[i]==‘[‘&&s[j]==‘]‘)d[i][j]=min(d[i][j],d[i+1][j-1]); for(int k=i;k<j;k++){ d[i][j]=min(d[i][j],d[i][k]+d[k+1][j]); } } } } int main(){ while(cin>>s){ n=s.length(); if(s=="end")break; dp(); cout<<n-d[0][n-1]<<endl; } return 0; }
原文：http://www.cnblogs.com/demodemo/p/4732586.html
