Brackets Sequence - POJ 1141 dp

最新推荐文章于 2021-09-13 20:44:00 发布

提比-我有特殊的AC技巧

最新推荐文章于 2021-09-13 20:44:00 发布

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分类专栏： dp动态规划和递推 POJ

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本文探讨了如何通过添加最少的符号来使给定的括号序列成为有效的括号序列，并提供了实现这一目标的算法。通过动态规划的方法，文章详细解释了如何计算并构建所需的最短有效括号序列。

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Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24740		Accepted: 6964		Special Judge

Description

Let us define a regular brackets sequence in the following way:

1. Empty sequence is a regular sequence.
2. If S is a regular sequence, then (S) and [S] are both regular sequences.
3. If A and B are regular sequences, then AB is a regular sequence.

For example, all of the following sequences of characters are regular brackets sequences:

(), [], (()), ([]), ()[], ()[()]

And all of the following character sequences are not:

(, [, ), )(, ([)], ([(]

Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

Input

The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output

Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

Sample Input

([(]

Sample Output

()[()]

题意：添加最少的符号，使得它是完美的。

思路：用dp[i][j]表示修i到j需要的最少符号数量。具体的看代码吧。

AC代码如下;

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[110];
int dp[110][110];
void solve(int l,int r)
{ if(l>r)
   return;
  if(l==r)
  { if(s[l]=='(' || s[l]==')')
     printf("()");
    else
     printf("[]");
    return;
  }
  if(dp[l][r]==dp[l+1][r-1] && s[l]=='(' && s[r]==')')
  { printf("(");
    solve(l+1,r-1);
    printf(")");
    return;
  }
  if(dp[l][r]==dp[l+1][r-1] && s[l]=='[' && s[r]==']')
  { printf("[");
    solve(l+1,r-1);
    printf("]");
    return;
  }
  int i,pos,ans=1000;
  for(i=l;i<r;i++)
   if(dp[l][i]+dp[i+1][r]<ans)
   { ans=dp[l][i]+dp[i+1][r];
      pos=i;
   }
  solve(l,pos);
  solve(pos+1,r);
}
int main()
{ int i,j,k,n;
  gets(s+1);
  n=strlen(s+1);
  for(i=1;i<=n;i++)
   dp[i][i]=1;
  for(i=n-1;i>=1;i--)
   for(j=i+1;j<=n;j++)
   { dp[i][j]=1000;
     if((s[i]=='(' && s[j]==')') || (s[i]=='[' && s[j]==']'))
     dp[i][j]=dp[i+1][j-1];
     for(k=i;k<j;k++)
      dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
   }
  solve(1,n);
  printf("\n");
}