POJ

Brackets

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

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题目大意：
找一个括号序列中，最大的匹配子串。

解题思路：
用f[i][j]表示 从第i个位置到第j个位置的最大匹配数。
1.如果边界两个匹配，那么f[i][j] = f[i+1][j-1]+2。
2.如果边界不匹配，那么f[i][j] = max(f[i][j] , f[i][x] + f[x+1][j])。

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源代码：

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;int n;char a[1005];int f[1005][1005];int p,q;int main(){    while(scanf("%s",a)){        if(a[0]=='e')          break;        n = strlen(a);        memset(f,0,sizeof(f));        for(int len=1;len<n;len++){            for(int i=0,j=len;j<n;i++,j++){                if(a[i]=='('&&a[j]==')' || a[i]=='['&&a[j]==']')                     f[i][j] = f[i+1][j-1]+2;                for(int x=i;x<j;x++)                       f[i][j]=max(f[i][j],f[i][x]+f[x+1][j]);                //cout<<i<<"  "<<j<<"   "<<f[i][j]<<endl;            }        }        printf("%d\n",f[0][n-1]);    }    return 0;}