Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14226 Accepted: 7476
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 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
Source

Stanford Local 2004

#include<algorithm> #include<iostream> #include<cmath> #include<cstring> #include<cstdio> using namespace std; bool match(char a,char b); #define mst(a,b) memset((a),(b),sizeof(a)) const int maxn=500; int dp[maxn][maxn]; int main() {string ob;while(cin>>ob){if(ob=="end") break;mst(dp,0);for(int len=2;len<=ob.length( );len++){for(int i=1;i<=ob.length( )+1-len;i++){int j=len+i-1;if(match(ob[i-1],ob[j-1])) 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]);}}}// for(int len=1;len<ob.length( )-2;len++) cout<<dp[len][len+3]<<' ';cout<<dp[1][ob.length( )]<<endl;} } bool match(char a,char b) {if(a=='('&&b==')') return 1;if(a=='['&&b==']') return 1;else return 0; }

總結

以上是生活随笔為你收集整理的POJ 2955 区间DP必看的括号匹配问题，经典例题的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。