POJ 2955 Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2938		Accepted: 1516

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 … 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 [([])].

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

Stanford Local 2004

 

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 long

10 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 }

View Code