poj2955Brackets(区间DP)

最新推荐文章于 2019-07-10 00:08:39 发布

转载最新推荐文章于 2019-07-10 00:08:39 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/gcczhongduan/p/5058670.html

本文介绍了一种使用动态规划算法解决寻找字符串中最长的有效括号子序列的问题，并提供了一个具体的C语言实现示例。

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

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\n",dp[0][len-1]);
    }
}

转载于:https://www.cnblogs.com/gcczhongduan/p/5058670.html