poj2955:Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8716		Accepted: 4660

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

6
6
4
0
6

分析

dp[l][r]表示区间[l,r]的答案。

状态转移方程，详见代码

• dp[i][j] = max(dp[i+1][j],dp[i][j-1]);
• if ((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
  　　dp[i][j]=dp[i+1][j-1]+2;
• dp[l][r]=max(dp[l][r],dp[l][k]+dp[k+1][r]);

code

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[110];
int dp[110][110];

int main()
{
    while (scanf("%s",s)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        if (s[0]=='e') break;
        int len = strlen(s);
        for (int i=len-2; i>=0; --i)
        {
            for (int j=i; j<len; ++j)
            {
                dp[i][j] = max(dp[i+1][j],dp[i][j-1]);
                if ((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                    dp[i][j]=dp[i+1][j-1]+2;
                for (int k=i; k<j; ++k) 
                    dp[i][j] = max(dp[i][j],dp[i][k]+dp[k+1][j]);
            }
        }
        printf("%d
",dp[0][len-1]);
    }
    return 0;
}