求最长公共子序列

"In biological applications , we often want to compare the DNA of two (or more) different organisms.....
 For example, the DNA of one organism may be
S1=  { ACCGGTCGAGTGCGCGGAAGCCGGCCGAA}
S2= {GTCGTTCGGAATGCCGTTGCTCTGTAAA}
one goal of comparing two strands of DNA is to determine how "similar" the two strands are, as some measure of how closely related the two organisms are."
------------------------------------------摘自Introduction to algorithms[算法导论]

现在,我们遇到求公共子序列的时候,公共子序列意味着什么?匹配?相似度?怎么求最长公共子序列(longest common subsequence)?本文将告诉你其算法并给出c/c++ 与Java实现源代码.

[1] 定义:何为最长公共子序列?
  eg:  X = {A,B,C,B,D,A,B}   Y = {B,D,C,A,B,A}此两序列的最长公共子序列是LCS={B,C,B,A}
  定义我就不多说了,自己感受咯.

[2] 一个相关的定理:
     若X = {x1,x2,...,xm} , Y = {y1,y2,...,yn}的LCS是Z={z1,z2,...,zk}那么有:
      1) 如果xm = yn ,  则zk = xm = yn 并且 Zk-1  是Xm-1和Yn-1的LCS.
      2)如果xm ≠ yn, 那么zk ≠ xm意味着Z 是Xm-1和Y的LCS.
      3)如果xm ≠ yn, 那么zk ≠ yn 意味着Z是X和Yn-1的LCS.
这个定理其实不难理解,用反证法可以证明之.但是却暗含递归思想.

let us define c[i,j] to be the length of an LCS of the sequence Xi and Yj.
整理一下.有如下的递归关系:

[3]算法源程序

#include  < stdio.h >
#include  < string .h >

  int *  LCS_length( char *  X,  char *  Y)
{
     int  m  =  strlen(X)  +   1 ;
     int  n  =  strlen(Y)  +   1 ;
    
     int  ( * c)[n]  =   new   int [m][n];

     int  ( * b)[n]  =   new   int [m][n];
     for ( int  i = 0 ;i < m; ++ i)
         for ( int  j = 0 ;j < n; ++ j)
           b[i][j]  =   0 ;
       
     for ( int  i = 0 ;i < m; ++ i)
        c[i][ 0 ]  =   0 ;
     for ( int  i = 0 ;i < n; ++ i)
        c[ 0 ][i]  =   0 ;
        
     for ( int  i = 1 ;i < m; ++ i)
         for ( int  j = 1 ;j < n; ++ j)
        {
                 if  (X[i - 1 ]  ==  Y[j - 1 ])
                {
                    c[i][j]  =  c[i - 1 ][j - 1 ]  + 1  ;
                    b[i][j]  =   2 ;   //  比较相等之标记 
                }    
                 else    
                {
                     if ( c[i - 1 ][j]  >=  c[i][j - 1 ])
                    {
                         c[i][j]  =  c[i - 1 ][j];
                         b[i][j]  =   1 ;  //
                     }    
                     else
                         c[i][j]  =  c[i][j - 1 ];
                 }       
        }         
      
    delete[] c;
    
     return  ( int * )b;
}    

void  LCS( char *  X, char *  Y, int  m,  int  n, int *  b)
{   
     if (m  == 0   ||  n  ==   0 )
         return ;
     if (b[m *  (strlen(Y) + 1 ) +  n]  ==   2 )
    {
        LCS(X,Y,m - 1 ,n - 1 ,b);
        printf( " %c " ,X[m - 1 ]);
    }
     else   if (b[m *  (strlen(Y) + 1 ) +  n]  ==   1 )    
    {
        LCS(X,Y,m - 1 ,n,b);
    }    
     else
    {
        LCS(X,Y,m,n - 1 ,b);
    }      
}

int  main()
{
    //  char* X = "ABCBDAB";
    //  char* Y = "BDCABA";
     char *  X  =   " ACCGGTCGAGTGCGCGGAAGCCGGCCGAA " ; 
     char *  Y  =   " GTCGTTCGGAATGCCGTTGCTCTGTAAA " ;
     int  m  =  strlen(X)  +   1 ;
     int  n  =  strlen(Y)  +   1 ;
   
     int *  c  =  NULL;
    c   =  LCS_length(X,Y);
 
    LCS(X,Y,strlen(X),strlen(Y),c);
    
    delete[] c;
    getchar();
     return   0 ;   
}

以上程序在VC6.0下是不能正常编译的.推荐用Dev-C++来调试.我用的版本是4.9.9.0.以上程序或许有点费解,主要是C++对动态的多维数组不支持! 您看,我到求LCS时变二维为一维来求了,但愿你能根据前面的定理看懂我这糟糕的程序咯.

那么Java怎么样?JAVA支持多维数组啦,因此java写起来"好看"多了.^_^

class  TestLCS
{ 
     final   static   int  NorthWest  =   2 ;
     final   static   int  UP  =   1 ;
     final   static   int  LEFT  =   0 ;
     private   static   int [][] LCS_length(String X, String Y)
    {
         int  m  =  X.length() + 1 ;
         int  n  =  Y.length() + 1 ;
        
         int [][] c  =   new   int [m][n];
         int [][] flag  =   new   int [m][n];
        
         for ( int  i = 0 ;i < m; ++ i)
             for ( int  j = 0 ;j < n; ++ j)
                flag[i][j]  =  LEFT;
        

         for ( int  i = 0 ;i < m; ++ i)    // 递归表达式2
            c[i][ 0 ]  =   0 ;
         for  ( int  i = 0 ;i < n; ++ i)
            c[ 0 ][i]  =   0 ;
    
        
         for ( int  i = 1 ;i < m; ++ i)
             for ( int  j = 1 ;j < n; ++ j)
            {
                 if ( X.charAt(i - 1 )  ==  Y.charAt(j - 1 ))
                {
                    c[i][j]  =  c[i - 1 ][j - 1 ]  + 1 ; // 递归表达式3
                    flag[i][j]  =  NorthWest;     //
                }
                 else
                {
                     if (c[i - 1 ][j]  >  c[i][j - 1 ])
                    {
                        c[i][j]  =  c[i - 1 ][j];
                        flag[i][j]  =  UP;  //
                    }
                     else
                    {
                        c[i][j]  =  c[i][j - 1 ];
                    }
                }
                    
            }
        
         return  flag;
    }

     private   static   void  LCS(String X,  int  m, int  n, int [][] flag)
    {
         if (m  == 0   ||  n  == 0 )  return ;
        
         if (flag[m][n]  == NorthWest)
        {
            LCS(X,m - 1 ,n - 1 ,flag);
            System.out.print(X.charAt(m - 1 ));
        }
         else   if (flag[m][n]  ==  UP)
        {
            LCS(X,m - 1 ,n,flag);
        }
         else
        {
            LCS(X,m,n - 1 ,flag);
        }
    }
     public   static   void  main(String[] args) 
    {
             // String X = "ABCBDAB";
             // String Y = "BDCABA";
            String X  =   " ACCGGTCGAGTGCGCGGAAGCCGGCCGAA " ;
            String Y  =   " GTCGTTCGGAATGCCGTTGCTCTGTAAA " ;
            
             int  m  =  X.length();
             int  n  =  Y.length();
            
             int [][] flag  =   new   int [m + 1 ][n + 1 ];
            
            flag   =  LCS_length(X,Y);
            LCS(X,m,n,flag);
        
             // 程序输出:GTCGTCGGAAGCCGGCCGAA
    }

}

 本文完.如有问题欢迎留言讨论.

参考资料:<Introduction to algorithms>