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 msuch 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

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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

•题目大意：给出一个括号序列，求出其中括号最大匹配数量

•第一步：确定状态

•f[i][j] 表示ai……aj的串中，有多少个已经匹配的括号

•第二步：确定状态转移方程

•如果ai与aj是匹配的

•f[i][j] = f[i + 1][j - 1] + 2

•否则f[i][j] = max(f[i][j],f[i][k] + f[k + 1][j])

•（相当于是将i到j分成[xxxxx]xxxxx两部分）

•边界 f[i][i] = 0

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=150;
int dp[maxn][maxn];//表示字符串s的第i..j字符需要最少括号数
char a[maxn];
bool cmp(int n,int m)
{
    if((n=='('&&m==')')||(n=='['&&m==']'))
        return 1;
    return 0;
}
int main()
{
    //while(scanf("%s",a)!=EOF){
    while(scanf("%s",a+1)!=EOF&&a[1]!='e'){//下标从1开始
        memset(dp,0,sizeof(dp));
        int n=strlen(a+1);
        for(int len=2;len<=n;len++)
        {
            for(int i=1;i<=n;i++)
            {
                int j=i+len-1;
                if(j>n) break;

                if(cmp(a[i],a[j]))//如果ai与aj是匹配的
                {
                    if(i+1>j-1) dp[i][j]=2;//中间长度为0
                    else dp[i][j]=dp[i+1][j-1]+2;
                }
                for(int k=i;k<j;k++)//中断点
                {
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
            }
        }
        printf("%d\n",dp[1][n]);
    }
    return 0;
}