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poj -2955 Brackets

2024-07-03 23:59:28   229人阅读

http://poj.org/problem?id=2955

Brackets
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2707   Accepted: 1403

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 msuch that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … 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
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#include<stdio.h>
#include<string.h>
int dp[100][100];
int max(int x,int y)
{
    if(x>y)
       return x;
    else
      return y;
}
bool match(char x,char y)
{
    if(x==‘[‘&&y==‘]‘)
       return true;
    else if(x==‘(‘&&y==‘)‘)
       return true;
    else
        return false;
}
int main()
{
      int len,i,j,k,g;
     char str[100];
     while(gets(str))
     {
         if(str[0]==‘e‘)
           break;
        memset(dp,0,sizeof(dp));
         len=strlen(str);
         for(i=0;i<len;i++)
             {
                 dp[i][i]=0;
                 if(match(str[i],str[i+1])) dp[i][i+1]=2;//初始值。
             }
 
         for(k=0;k<len;k++)
              for(i=0;i<len-k;i++)
              {
                  j=i+k;
                  if(match(str[i],str[j])) dp[i][j]=dp[i+1][j-1]+2;
                  for(g=0;g<k;g++)
                     dp[i][j]=max(dp[i][j],dp[i][i+g]+dp[i+g+1][j]);
              }
        printf("%d\n",dp[0][len-1]);
    }
    return 0;
}

 别人的代码，可供参考。

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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1002;
char s[maxn];
int dp[maxn][maxn];
int main()
{
    //freopen("//media/学习/ACM/input.txt","r",stdin);
    while(scanf("%s",s),s[0]!=‘e‘)
    {
        int i,j,k,n=strlen(s);
        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((s[i]==‘(‘&&s[k]==‘)‘)||(s[i]==‘[‘&&s[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]);
                 }
               //  cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
            }
        }
        cout<<dp[0][n-1]<<endl;
    }
    return 0;
}

 


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