POJ2955(区间DP)

Posted 2020-06-16

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5083		Accepted: 2733

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

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char buf[105];
int dp[105][105];
int main()
{
    while(scanf("%s",buf)!=EOF&&buf[0]!=‘e‘)
    {
        memset(dp,0,sizeof(dp));
        int len=strlen(buf);
        
        for(int l=1;l<len;l++)//区间长度
        {
            for(int i=0,j=l;j<len;i++,j++)
            {
                if((buf[i]==‘(‘&&buf[j]==‘)‘)||(buf[i]==‘[‘&&buf[j]==‘]‘))
                    dp[i][j]=dp[i+1][j-1]+2;
                for(int k=i+1;k<=j-1;k++)//"()[]()" dp[i+1][j-1]=2 而dp[i][j]=6
                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]);
            }    
        }
        printf("%d\n",dp[0][len-1]);
    }
    
    return 0;
}

一种普适的字符串DP顺序

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;char buf[105];
int dp[105][105];
int main()
{
    while(scanf("%s",buf)&&buf[0]!=‘e‘)
    {
        memset(dp,0,sizeof(dp));
        int len=strlen(buf);
        
        for(int i=0;i<len;i++)
        {
            for(int j=i-1;j>=0;j--)
            {
                if((buf[j]==‘(‘&&buf[i]==‘)‘)||(buf[j]==‘[‘&&buf[i]==‘]‘))
                    dp[j][i]=dp[j+1][i-1]+2;
                for(int x=j+1;x<i;x++)
                    dp[j][i]=max(dp[j][i],dp[j][x]+dp[x][i]);
            }
        }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}