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POJ 2955 Brackets

2024-08-15 22:21:29   222人阅读

Brackets

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 6622 Accepted: 3558

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

66406

Source

Stanford Local 2004
 
 
题目大意：给你一个长度不超过100的括号序列，求最长合法括号子序列的长度。合法的括号序列满足下列条件：
1.空的括号序列是合法的；
2.如果一个括号序列s是合法的，那么(s)和[s]都是合法的；
3.如果两个括号序列a和b都是合法的，那么ab也是合法的；
4.其他的括号序列都是不合法的。
例如：(), [], (()), ()[], ()[()]都是合法的，而(, ], )(, ([)], ([(]都是不合法的。
 
解题思路：一道典型的区间DP模型题目。分析一下问题，可以发现：如果找到一对匹配的括号，例如[xxx]oooo，就把区间分成两部分，一部分是xxx，另一部分是oooo。
设dp[i][j]表示区间[i,j]之间的最长合法括号子序列的长度，那么当i<j时，如果区间[i+1,j]内没有与i匹配的括号，则dp[i][j]=dp[i+1][j]；如果存在一个与之匹配的k，那么dp[i][j]=max{dp[i+1][j], dp[i+1][k-1]+dp[k+1][j]+1(i<=k<=j&&i和k是一对匹配的括号)}。因此，我们将整个串长作为区间进行搜索，那么最后2*dp[0][len-1]即为答案，len表示串的长度。详见代码。
 
 
附上AC代码：
技术分享
 1 #include <cstdio> 2 #include <cstring> 3 using namespace std; 4 const int maxn = 105; 5 char str[maxn]; 6 int dp[maxn][maxn]; 7  8 bool match(char a, char b){ 9     return (a==‘(‘&&b==‘)‘) || (a==‘[‘&&b==‘]‘);10 }11 12 int dfs(int l, int r){13     if (l >= r)14         return 0;15     if (l+1 == r)16         return dp[l][r] = match(str[l], str[r]);17     if (dp[l][r])18         return dp[l][r];19     int ans = dfs(l+1, r);20     for (int i=l; i<=r; ++i)21         if (match(str[l], str[i])){22             int t = dfs(l+1, i-1)+dfs(i+1, r)+1;23             if (t > ans)24                 ans = t;25         }26     return dp[l][r] = ans;27 }28 29 int main(){30     while (~scanf("%s", str) && str[0]!=‘e‘){31         memset(dp, 0, sizeof(dp));32         int len = strlen(str);33         dfs(0, len-1);34         printf("%d\n", 2*dp[0][len-1]);35     }36     return 0;37 }
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POJ 2955 Brackets


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