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POJ2955 Brackets (区间DP)

时间:2020-03-18 13:48:36      阅读:54      评论:0      收藏:0      [点我收藏+]

标签:name   cout   ons   pre   mem   word   sub   -o   amp   

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 a1a2an, 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, …, im where 1 ≤ i1 < i2 < … < imn, ai1ai2aim 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
区间DP经典例题。因为括号是可以嵌套并列的,所以想到区间DP(才不是因为看到了标签..这个DP数组的含义比较好想到,dp[i][j]表示下标i~j的这一段字符串最多有多少个括号匹配了,按照区间DP的常见套路,第一层枚举区间长度,由于匹配最少需要两个括号,所以len从2开始;第二层枚举左端点,同时也可以根据左端点和长度确定右端点。这里先判断一步,如果s[i]和s[j]匹配了,那么dp[i][j]起码是dp[i+1][j-1]+2(为什么说是起码呢?因为考虑样例的()()()这个序列自然就明白了)
然后找决策点,看看由两个分开的子区间能不能转移到当前并且使得答案更大。最后输出的就是dp[0][len-1]。
#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;
char s[105];
int dp[105][105];
bool check(int l,int r)
{
    if(s[l]==(&&s[r]==))return true;
    if(s[l]==[&&s[r]==])return true;
    return false;
}
int main()
{
    while(scanf("%s",s)&&s[0]!=e)
    {
        memset(dp,0,sizeof(dp));
        int len,i,j,k;
        for(len=2;len<=strlen(s);len++)
        {
            for(i=0;i+len-1<=strlen(s)-1;i++)
            {
                int j=i+len-1;
                if(check(i,j))
                {
                    dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
                }
                for(k=i;k<=j-1;k++)
                {
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                }
                
            }
        }
        cout<<dp[0][strlen(s)-1]<<endl;
    }
    return 0;
}

 




POJ2955 Brackets (区间DP)

标签:name   cout   ons   pre   mem   word   sub   -o   amp   

原文地址:https://www.cnblogs.com/lipoicyclic/p/12516612.html

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