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 … a_n, 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, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

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

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.

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

解题：

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 const int maxn = 102;12 char str[maxn<<1];13 int dp[maxn][maxn];14 int main() {15     int i,j,k,len,t,ans;16     while(gets(str)&&strcmp(str,"end")) {17         len = strlen(str);18         memset(dp,0,sizeof(dp));19         ans = 0;20         for(i = 0; i < len; i++) {21             for(j = 0,k = i; k < len; k++,j++) {22                 if((str[j] == ‘(‘ && str[k] == ‘)‘) || (str[j] == ‘[‘ && str[k] == ‘]‘)) {23                     dp[j][k] = dp[j+1][k-1] + 2;24                 }25                 for(t = j+1; t < k; t++)26                     if(dp[j][t]+dp[t][k] > dp[j][k]) dp[j][k] = dp[j][t]+dp[t][k];27                 if(dp[j][k] > ans) ans = dp[j][k];28             }29 30         }31         printf("%d\n",ans);32     }33     return 0;34 }

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