一个粗糙的打印两序列所有最大公共子序列的方法

以下是典型动态算法打印两序列最大公共子序列的方法，该方法只能打印一个最大公共子序列：

 1 #include <iostream>
 2 using std::cout;
 3 using std::endl;
 4 
 5 const int NUM1=30,NUM2=30;
 6 char result[NUM1+1];
 7 int count=0;
 8 int lcsLength(char x[], char y[], int b[][NUM2+1],int m,int n) 
 9 {
10     int c[NUM1+1][NUM2+1];
11     for (int i = 0; i<=m; i++) 
12         c[i][0]=0;
13     for (int i = 0; i<=n; i++ ) 
14         c[0][i]=0;
15     for (int i = 1; i <= m; i++) 
16     {
17         for (int j = 1; j <= n; j++) 
18         {
19             if (x[i - 1]==y[j - 1])
20             {
21                 c[i][j]=c[i-1][j-1]+1; 
22                 b[i][j]=1;
23             }
24             else 
25                 if (c[i-1][j]>c[i][j-1]) 
26                 {
27                     c[i][j]=c[i-1][j]; 
28                     b[i][j]=2;
29                 }
30                 else 
31                 {
32                     c[i][j]=c[i][j-1]; 
33                     b[i][j]=3;
34                 }
35         }
36     }
37     return c[m][n];
38 }
39 
40 void lcs(int i,int j, char x[], int b[][NUM2+1] ) 
41 {
42     if (i ==0 || j==0) 
43         return;
44 
45     if(b[i][j]== 1) 
46     {    
47         lcs(i-1, j-1, x, b);
48         cout<<x[i-1];
49     }
50     else 
51         if(b[i][j] == 2)
52             lcs(i-1, j, x, b);
53         else 
54             lcs(i, j-1, x, b);
55 }
56 
57 int _tmain(int argc, _TCHAR* argv[])
58 {
59     char * x = "3123";
60     char * y ="132";
61     int xlen = strlen(x);
62     int ylen = strlen(y);
63     int b[NUM1 + 1][NUM2 + 1];
64     int len = lcsLength( x, y, b, xlen,ylen );
65     cout<<len<<endl;
66     lcs( xlen,ylen, x, b);
67 
68     getchar();
69     return 0;
70 }

只考虑最终的最优解，而不输出所有最优解或许是动态算法的很大的缺点。就这个问题，下面是一个粗糙的打印两序列所有最大公共子序列的方法：

 1 #include <iostream>
 2 using std::cout;
 3 using std::endl;
 4 
 5 const int NUM1=30,NUM2=30;
 6 char result[NUM1+1];
 7 int count=0;
 8 int lcsLength(char x[], char y[], int b[][NUM2+1],int m,int n) 
 9 {
10     int c[NUM1+1][NUM2+1];
11     for (int i = 0; i<=m; i++) 
12         c[i][0]=0;
13     for (int i = 0; i<=n; i++ ) 
14         c[0][i]=0;
15     for (int i = 1; i <= m; i++) 
16     {
17         for (int j = 1; j <= n; j++) 
18         {
19             if (x[i - 1]==y[j - 1])
20             {
21                 c[i][j]=c[i-1][j-1]+1; 
22                 b[i][j]=1;
23             }
24             else 
25                 if (c[i-1][j]>c[i][j-1]) 
26                 {
27                     c[i][j]=c[i-1][j]; 
28                     b[i][j]=2;
29                 }
30                 else 
31                     if(c[i-1][j]<c[i][j-1])
32                     {
33                         c[i][j]=c[i][j-1]; 
34                         b[i][j]=3;
35                     }
36                     else
37                     {
38                         c[i][j]=c[i][j-1];    //或者c[i][j]=c[i-1][j];
39                         b[i][j]=4;
40                     }
41         }
42     }
43     return c[m][n];
44 }
45 
46 void lcs(int i,int j, char x[], int b[][NUM2+1],int idx,int len) 
47 {
48     if (i ==0 || j==0) 
49     {
50         for(int s=0;s<len;s++)
51             cout<<result[s];
52         cout<<endl;
53         count++;
54         return;
55     }
56     int t=b[i][j];
57     if(b[i][j]== 1) 
58     {    
59         idx--;
60         result[idx]=x[i- 1];
61         lcs(i-1, j-1, x, b,idx,len);
62     }
63     else 
64         if(b[i][j] == 2)
65             lcs(i-1, j, x, b,idx,len);
66         else 
67             if(b[i][j]==3)
68                 lcs(i, j-1, x, b,idx,len);
69             else
70             {
71                 lcs(i,j-1,x,b,idx,len);
72                 lcs(i-1,j,x,b,idx,len);
73             }
74 }
75 
76 int _tmain(int argc, _TCHAR* argv[])
77 {
78     char * x = "3123";
79     char * y ="132";
80     int xlen = strlen(x);
81     int ylen = strlen(y);
82     int b[NUM1 + 1][NUM2 + 1];
83     int len = lcsLength( x, y, b, xlen,ylen );
84     cout<<len<<endl;
85     lcs( xlen,ylen, x, b,len,len );
86     cout<<"共有："<<count<<"种";
87     getchar();
88     return 0;
89 }

posted on 2012-05-09 12:16  NLP新手  阅读(305)  评论(0)  编辑  收藏  举报

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