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POJ 2955-Brackets(区间DP)
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 3340 | Accepted: 1716 |
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, …, imwhere 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
经典区间DP。
题意:求括号匹配数。
dp[l][r]为区间[l,r]内括号匹配数,dp[l][r]=max(dp[l][r],dp[l+1][k-1]+dp[k+1][r]+2)(s[l]和s[k]匹配)
枚举区间,形式大都一样,3层for循环,看到很多人说区间DP写成记忆化搜索比较容易懂。。orz
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cctype> #include <cmath> #include <cstdlib> #include <vector> #include <queue> #include <set> #include <map> #include <list> #define ll __int64 #define pp pair<int,int> using namespace std; const int INF = 0x3f3f3f3f; char s[110]; int dp[110][110]; void solve() { memset(dp, 0, sizeof(dp)); int len = strlen(s) - 1; for (int p = 1; p <= len; p++) { for (int l = 1; l <= len; l++) { int r = l + p - 1; if (r > len) { break; } dp[l][r] = dp[l + 1][r]; for (int k = l + 1; k <= r; k++) if ((s[l] == '(' && s[k] == ')') || (s[l] == '[' && s[k] == ']')) { dp[l][r] = max(dp[l][r], dp[l + 1][k - 1] + dp[k + 1][r] + 2); } } } printf("%d\n", dp[1][len]); } int main() { while (~scanf("%s", s + 1) && s[1] != 'e') { s[0] = 2; solve(); } return 0; }
POJ 2955-Brackets(区间DP)