POJ 2955 Brackets

最新推荐文章于 2019-08-24 09:36:00 发布

转载最新推荐文章于 2019-08-24 09:36:00 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/zjdx1998/p/4049158.html

本文探讨了如何通过动态规划解决寻找给定字符序列中包含的最长合法括号子序列的问题，详细解释了算法原理并提供了代码实现。

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Brackets

Time Limit: 1000MS		Memory Limit: 65536K

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

继续刷水，和上一题差不多,dp[i][j]=max(dp[i+1][j-1]+1//i,j匹配,dp[i][k]+dp[k+1][j]);

 1 #include<set>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 const int N = 110;
 9 #define For(i,n) for(int i=1;i<=n;i++)
10 #define Rep(i,l,r) for(int i=l;i<=r;i++)
11 #define Down(i,r,l) for(int i=r;i>=l;i--)
12 
13 char s[N];
14 int dp[N][N],n;
15 //dp[i][j]=max{dp[i+1][j-1]+1,dp[i][k]+dp[k+1][j]}
16 
17 bool match(char A,char B){
18     if(A=='(') return (B==')');
19     if(A=='[') return (B==']');
20     return false;
21 }
22 
23 void DP(){
24     memset(dp,0,sizeof(dp));
25     n=strlen(s+1);
26     Down(i,n-1,1)
27       Rep(j,i+1,n){
28           if(match(s[i],s[j])) dp[i][j]=dp[i+1][j-1]+1;
29           Rep(k,i,j-1)
30             dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
31       }
32     cout<<dp[1][n]*2<<endl;
33 }
34 
35 int main(){
36     while(scanf("%s",s+1),strcmp(s+1,"end")){
37         DP();
38     }
39     return 0;
40 }