区间DP POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 29520		Accepted: 8406		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

()[()]
　　　　

/*对于DP题目需要记录到底是怎样得到结果的，不一定了可以通过记录信息直接得到，可以选择数组记录其他有用信息，可以写递归来找出最后的答案。*/
#define N 111
#include<iostream>
using namespace std;
#include<cstring>
#include<cstdio>
#define inf (1<<31)-1
int f[N][N],pos[N][N],lens;
char s[N];
void print(int l,int r)/*输出序列的过程*/
{
    if(l<=r)/*一定要有这一句，否则对于相邻的‘（）’，就会死循环了*/
    {
        if(l==r)/*能到这一步，说明只能补上括号了*/
        {
            if(s[l]=='('||s[l]==')') printf("()");
            if(s[l]=='['||s[l]==']') printf("[]");
        }
        else 
        {
            if(pos[l][r]==-1)/*说明该区间最左括号与最右匹配*/
            {
                printf("%c",s[l]);
                print(l+1,r-1);/**/
                printf("%c",s[r]);
            }
            else 
            {
                print(l,pos[l][r]);
                print(pos[l][r]+1,r);
            }
        }
    }
}
int main()
{
//    freopen("bracket.in","r",stdin);
//    freopen("bracket.out","w",stdout);
    scanf("%s",s+1);
    lens=strlen(s+1);
    for(int i=1;i<=lens;++i)
      f[i][i]=1;
/*    for(int i=lens-1;i>=1;--i)
      for(int j=i+1;j<=lens;++j)
      {
            f[i][j]=inf;
            if(((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')))
              {
                  f[i][j]=f[i+1][j-1];
                  pos[i][j]=-1;
            }
                                
            for(int k=i;k<=j-1;++k)
            {
            if(f[i][j]>f[i][k]+f[k+1][j])
            {
                    f[i][j]=f[i][k]+f[k+1][j];
                    pos[i][j]=k;
            }
            
          }
          
      }这两种都是可以得出正确答案的，但是我建议使用下面的，对于区间DP，最外层循环最好枚举区间长度，内层枚举区间*/
    for(int k=1;k<lens;++k)
      for(int i=1,j=i+k;j<=lens&&i<=lens;++j,++i)
      {
            f[i][j]=inf;
            if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
              {
                  f[i][j]=f[i+1][j-1];
                  pos[i][j]=-1;
            }
                 /*不要加else，因为即使当前区间的最左和最右匹配，也不一定比放弃他们匹配优*/
            for(int k=i;k<=j-1;++k)
            {
            if(f[i][j]>f[i][k]+f[k+1][j])
            {
                    f[i][j]=f[i][k]+f[k+1][j];
                    pos[i][j]=k;
            }
            
          }
          
      }
    print(1,lens);
    printf("
");/*坑爹的POJ，没有这句，一直没对*/
    //fclose(stdin);fclose(stdout);
    return 0;
}