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http://poj.org/problem?id=2955
Description
We give the following inductive definition of 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
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; const int INF = 0x3f3f3f3f; #define N 105 char s[N]; int dp[N][N]; int main() { while(scanf("%s", s), strcmp(s, "end")) { int i, j, k, len=strlen(s)-1; memset(dp, 0, sizeof(dp)); for(i=len-1; i>=0; i--) { for(j=i+1; j<=len; j++) { dp[i][j] = dp[i+1][j]; for(k=i+1; k<=j; k++) { if((s[i]==‘(‘ && s[k]==‘)‘) || (s[i]==‘[‘ && s[k]==‘]‘)) { if(k==i+1) dp[i][j] = max(dp[i][j], dp[k+1][j]+2); else dp[i][j] = max(dp[i][j], dp[i+1][k-1]+dp[k+1][j]+2); } } } } printf("%d\n", dp[0][len]); } return 0; }
(区间dp 或 记忆化搜素 )Brackets -- POJ -- 2955
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原文地址：http://www.cnblogs.com/YY56/p/5051617.html
