Brackets（区间dp）

Brackets

Time Limit: 1000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name:Main

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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:

<code>(), [], (()), ()[], ()[()]

while the following character sequences are not:

<code>(, ], )(, ([)], ([(]

Given a brackets sequence of characters a₁a₂ …a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence ofs. That is, you wish to find the largest m such that for indicesi₁, i₂, …, i_m where 1 ≤i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 …a_im is a regular brackets sequence.

Given the initial sequence <code>([([]])], the longest regular brackets subsequence is <code>[([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters <code>(, <code>), <code>[, and <code>]; 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

66406

跟 NYOJ 15 括号匹配（二）（区间dp）只是要求反了一下。。。这里，dp[i][j] 代表从i到j的范围， 最大的括号匹配数。如果a[i] == a[j] 那么，dp[i][j] = dp[i+1][j-1] + 2 ,然后，进行长度区间值更新。。

#include <iostream>#include <string.h>#include <algorithm>using namespace std;int dp[500][500];int main(){    char a[500];    while(cin >> a && a[0] != 'e'){        int len = strlen(a);        memset(dp, 0, sizeof(dp));        for(int l = 1; l < len; l++)            for(int i = 0; i < len - 1; i++){//线段长度递推                int j = i + l;                if((a[i]=='[' && a[j]==']') || (a[i]=='(' && a[j]==')'))                    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]);            }        cout << dp[0][len-1] << endl;    }    return 0;}