POJ 2955 Brackets（动态规划）_综合

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

【题解】 http://www.adapter1.cn/post/69061863001966917059.html

【AC代码】

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=200;
int dp[N][N];
char s[N];char Get(char c)
{if(c==']') return '[';if(c==')') return '(';return '!';
}int max(int x,int y)
{return x>y?x:y;
}void solve()
{memset(dp,0,sizeof(dp));while(~scanf("%s",s+1)&&s[1]!='e'){s[0]='?';int t=strlen(s)-1;for(int len=1;len<t;++len){for(int i=1;len+i<=t;++i){int j=i+len;char ch = Get(s[j]);dp[i][j]=dp[i][j-1];//更新当前最大值for(int k=i;k<j;++k)if(s[k]==ch){dp[i][j]=max(dp[i][j],dp[i][k-1]+dp[k+1][j-1]+2);}}}printf("%d\n",dp[1][t]);}
}int main()
{solve();return 0;
}