POJ 2955 Brackets (區間dp 括號匹配)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3951 | Accepted: 2078 |
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
- 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
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題目鏈接:
id=2955">http://poj.org/problem?
id=2955
題目大意:給一個括號序列,問序列中合法的括號最多有多少個。若A合法。則[A],(A)均合法,若A,B合法則AB也合法
題目分析:和POJ 1141那道經典括號匹配類似,這題更簡單一些。想辦法把問題轉化,既然要求最大的括號匹配數,我們考慮加最少的括號。使得整個序列合法,這樣就轉變成1141那題。開下腦動類比二分圖最大匹配的性質,最大匹配+最大獨立集=點數,顯然要增加最少的點使序列合法,則加的最少的點數即為|最大獨立集|。我們要求的是原序列的|最大匹配|,以上純屬yy,以下給出轉移方程,和1141一模一樣
dp[i][i] = 1;
然後枚舉區間長度
1)外圍匹配:dp[i][j] = dp[i + 1][j - 1];
2)外圍不匹配。枚舉切割點:dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); (i <= k < j)
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int const INF = 0x3fffffff; char s[205]; int dp[205][205]; int main() { while(scanf("%s", s) != EOF && strcmp(s, "end") != 0) { int len = strlen(s); memset(dp, 0, sizeof(dp)); for(int i = 0; i < len; i++) dp[i][i] = 1; for(int l = 1; l < len; l++) { for(int i = 0; i < len - l; i++) { int j = i + l; dp[i][j] = INF; if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) dp[i][j] = dp[i + 1][j - 1]; for(int k = i; k < j; k++) dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); } } printf("%d\n", len - dp[0][len - 1]); } }
POJ 2955 Brackets (區間dp 括號匹配)