POJ 2955 Brackets，poj2955brackets
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
6
6
4
0
6
Source
Stanford Local 2004
題目大意：給你一個長度不超過100的括號序列，求最長合法括號子序列的長度。合法的括號序列滿足下列條件：
1.空的括號序列是合法的；
2.如果一個括號序列s是合法的，那麼(s)和[s]都是合法的；
3.如果兩個括號序列a和b都是合法的，那麼ab也是合法的；
4.其他的括號序列都是不合法的。
例如：(), [], (()), ()[], ()[()]都是合法的，而(, ], )(, ([)], ([(]都是不合法的。
解題思路：一道典型的區間DP模型題目。分析一下問題，可以發現：如果找到一對匹配的括號，例如[xxx]oooo，就把區間分成兩部分，一部分是xxx，另一部分是oooo。
設dp[i][j]表示區間[i,j]之間的最長合法括號子序列的長度，那麼當i<j時，如果區間[i+1,j]內沒有與i匹配的括號，則dp[i][j]=dp[i+1][j]；如果存在一個與之匹配的k，那麼dp[i][j]=max{dp[i+1][j], dp[i+1][k-1]+dp[k+1][j]+1(i<=k<=j&&i和k是一對匹配的括號)}。因此，我們將整個串長作為區間進行搜索，那麼最後2*dp[0][len-1]即為答案，len表示串的長度。詳見代碼。
附上AC代碼：
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1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 using namespace std;
5 const int maxn = 105;
6 char str[maxn];
7 int dp[maxn][maxn];
8
9 bool match(char a, char b){
10 return (a=='('&&b==')') || (a=='['&&b==']');
11 }
12
13 int dfs(int l, int r){
14 if (l > r)
15 return 0;
16 if (l == r)
17 return dp[l][r] = 0;
18 if (l+1 == r)
19 return dp[l][r] = match(str[l], str[r]);
20 if (dp[l][r] != -1)
21 return dp[l][r];
22 int ans = dfs(l+1, r);
23 for (int i=l; i<=r; ++i)
24 if (match(str[l], str[i]))
25 ans = max(ans, dfs(l+1, i-1)+dfs(i+1, r)+1);
26 return dp[l][r] = ans;
27 }
28
29 int main(){
30 while (~scanf("%s", str) && str[0]!='e'){
31 memset(dp, -1, 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 }
View Code
