POJ2955 Brackets （区间DP）

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 … 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, …, imwhere 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

题意：

给出一个字符串，寻找字符串中可以组成一对括号的有几个，可以组成一对结果+2。

思路很简单，典型的区间dp问题

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const INF=0x3fffffff;
char s[205];
int dp[205][205];
int main()
{
    while(~scanf("%s",s)&&strcmp(s,"end")!=0)
    {
        int len=strlen(s);
        memset(dp,0,sizeof(dp));
        for(int i=0;i<len;i++)
            dp[i][i]=1;
        for(int l=1;l<len;l++)
        {
            for(int i=0;i<len-l;i++)
            {
                int j=i+l;
                dp[i][j]=INF;
                if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
                 dp[i][j]=dp[i+1][j-1];
                for(int k=i;k<j;k++)
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
            }
        }
        printf("%d\n",len-dp[0][len-1]);
    }
    return 0;
}