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  • poj2955Brackets(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 i1i2, …, im where 1 ≤ i1 < i2 < … < im ≤ nai1ai2 … 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
    dp[i][j]表示在区间i到j匹配的最大数。

    #include<stdio.h>
    #include<string.h>
    int max(int a,int b)
    {
        return a>b?a:b;
    }
    int main()
    {
        int dp[105][105];
        char str[105];
        while(scanf("%s",str)>0&&strcmp(str,"end")!=0)
        {
            int len=strlen(str);
            memset(dp,0,sizeof(dp));
            for(int l=1;l<len;l++)//所求区间头尾相差长度
            for(int i=0;i<len-l;i++)//区间起始位置
            {
                int j=l+i;//区间尾部位置
                dp[i][j]=dp[i+1][j];//当第i个在这段区间内没有匹配的时
                for(int k=i+1;k<=j;k++)//当第i个与第k个位置匹配上时,状态转移例如以下
                if(str[i]=='('&&str[k]==')'||str[i]=='['&&str[k]==']')
                dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
            }
            printf("%d
    ",dp[0][len-1]);
        }
    }
    


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  • 原文地址:https://www.cnblogs.com/gcczhongduan/p/5058670.html
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