POJ 2955 DP动态规划

最新推荐文章于 2019-05-06 08:08:52 发布

iteye_1642

最新推荐文章于 2019-05-06 08:08:52 发布

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文章标签：数据结构与算法

本文介绍了一种算法，用于解决寻找给定括号序列中最长的有效括号子序列的问题。通过动态规划方法，该算法能高效地找出括号序列中最长的有效配对，并返回其长度。

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Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1659		Accepted: 831

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 … a_n, 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, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

Source

Stanford Local 2004

#include<stdio.h> #include<string.h> int dp[105][105]; char str[105]; bool match(char a,char b) { if(a=='('&&b==')') return true; if(a=='['&&b==']') return true; return false; } int main() { int i,j,k,l,t; while(scanf("%s",str)!=EOF) { if(strcmp(str,"end")==0) break; memset(dp,0,sizeof(dp)); l=strlen(str); for(k=2;k<=l;k++) { for(i=0;i+k-1<l;i++) { t=i+k-1; for(j=i;j<t;j++) { if(dp[i][t]<dp[i][j]+dp[j+1][t]) dp[i][t]=dp[i][j]+dp[j+1][t]; if(match(str[i],str[t])) { if(dp[i][t]<dp[i+1][t-1]+2) dp[i][t]=dp[i+1][t-1]+2; } } } } printf("%d/n",dp[0][l-1]); } return 0; }