ACM POJ 2955

A - Brackets

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2955

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

开始没想，直接匹配，，样例是过了，不过还是果断挖了

这是错误代码

#include <iostream>
#include <cstdio>
#include <cstring>


using namespace std;
char bracket[105];
bool visit[105];
int ans;

void Find(char c,int n,int l)
{
    for(int i = n+1; i<l; i++)
    {
        if(bracket[i] == c)
        {
            visit[i] =true;
            ans +=2;
            return;
        }
    }
    return;
}

int main()
{

    while(scanf("%s",&bracket))
    {
        ans = 0;
        if(bracket[0]=='e')
            return 0;
        memset(visit,0,sizeof(visit));
        int len = strlen(bracket);

        for(int i=0; i<len; i++ )
        {
            if(!visit[i])
            {
                if(bracket[i] == '(')
                    Find(')',i,len);

                if(bracket[i] == '[')
                    Find(']',i,len);

            }


        }
        printf("%d\n",ans);
    }
    return 0;
}

  ac code

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
char bracket[105];
int dp[105][105];
int ans;


int main()
{

    while(scanf("%s",&bracket),bracket[0]!='e')
    {
        int i,j,k,n=strlen(bracket);

        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                dp[i][j] = 0;
            }

        for(i=n-1; i>=0; i--)
            for(j=i+1; j<=n-1; j++)
            {
                dp[i][j] = max(dp[i+1][j],dp[i][j-1]);
                for(k=i+1; k<=j; k++)
                {
                    if((bracket[i]=='('&&bracket[k]==')')||(bracket[i]=='['&&bracket[k]==']'))
                        dp[i][k]=max(dp[i][k],dp[i+1][k-1]+2);

                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);

                }
            }
        printf("%d\n",dp[0][n-1]);
    }

    return 0;
}