POJ2955Brackets[区间DP]

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Brackets
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6585   Accepted: 3534

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 i1i2, …, im where 1 ≤ i1 < i2 < … < im ≤ nai1ai2 … 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


题意:求最大括号匹配

f[i][j]表示i到j的最大括号匹配
两种转移
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=105;
char s[N];
int f[N][N];
inline bool check(int i,int j){
    if(s[i]==[&&s[j]==]) return 1;
    if(s[i]==(&&s[j]==)) return 1;
    return 0;
}
int main(){
    while(~scanf("%s",s+1)){
        if(s[1]==e) break;
        memset(f,0,sizeof(f));
        int n=strlen(s+1);
        for(int i=n;i>=1;i--)
            for(int j=i+1;j<=n;j++){
                if(check(i,j)) f[i][j]=f[i+1][j-1]+2;
                for(int k=i;k<=j;k++) f[i][j]=max(f[i][j],f[i][k]+f[k][j]); 
            }
        printf("%d\n",f[1][n]);        
    }
}

 

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