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发表于 2012-11-22 11:13 |只看该作者 |倒序浏览

下面的过程是做序列的最佳局域匹配，我现在是想做序列的次最佳匹配并输出结果，程序编来很是困难。下面是我最佳匹配的C++代码,那位大神帮我修改下，做出次最佳匹配。perl代码和C++代码都可以。
1序列结构及数据
rps-0,B0393.1
I001CAATCAAATTTTTTTTAAGTTAAGTAAAAAGAACACTTAATAAATTATTTTAATATTGTTAATAGTAATGTC
E012CAATCTTAACTTCCAAATGCAGCAATACGTCTACAAGAGACGTTTTGACGGACCAAATATCATCAACGTCAAGAAGACCTGGGAGAAGCTTCTTCTCGCT
I002CATTCTGAAAAGATTTATCAACAAAAAAGATTAATAAATTTTAAATC
E034TCAAGGAGCCAAGACTTCTCGTCATCTCTGACCCACGTATCGATCATCAGGCTGTCACTGAGGCTTCCTACGTCGGAGTTCCAGTCATCTCCTTCGTCAA
I003CATATATAAATTAAATATTTTTTACATTTATACATAAACTTAACATC
E056CCGAGTCTCCACTCAAGTTGATCGACATCGGAGTTCCATGCAACAACAAGGGAGAACGCTCAATCGGACTTATGTGGTGGATGCTCGCTCGCGAGATCCT
rps-2
I001CATCAATAAAATAAGAATTTGGAAGTCTAAAACTTATAAATTTAAAGTC
E012GGACGTGGAGGACGTGGTGGCCGCGGAGGACGTGGTGGACGCGGAGGAAGAGCCGGACGCGGAGGAGAGAAGGAGACCGAATGGACTCCAGTCACCAAGC
I002CATTCATCAAATAAAAAAATCAAAATAATACCCCAATATTAAAGGTC
E034AACCTTAAAGACGAGGTCCTCAAGATCTCCCCAGTCCAGAAGCAGACCACTGCCGGACAACGTACCCGCTTCAAGGCTTTCGTTGCTATCGGAGATCACG

2 以下是我的C++代码
#include "stdio.h"
#include "stdlib.h"
#include "math.h"
#include "string.h"
#include"ctype.h"
#include"conio.h"

void main()
{

char  a[100000], b[100000], c[100000],d[5],d2[5], e[10],f[200],h[200],h1[200],f1[200],g[11]={ '0', '1', '2', '3', '4', '5', '6', '7', '8', '9','\0'};

long int i=0,j=0,k=0,k2=0,z1=0,z2=0,z3=0,m=0,m1=0,m2=0,m3=0,m4=0,n=0,n1=0,n2=0,n3=0,p=0,p1=0,q=0,q1=0;
FILE *fp1, *fp2;

for(i=0;i<200;i++)
{  h[i]='\0';
   f[i]='\0';
   h1[i]='\0';
   f1[i]='\0';
}
   fp1=fopen("序列结构数据.txt","r");
   fp2=fopen("局域比对结果.txt","w");





do
{

fgets(a,100000,fp1);

      if(a[0]=='@')
      break;
// fprintf(fp2,"%s", a);

if((a[0]!='I')&&(a[0]!='E')&&(a[0]!='S'))

   {

         fprintf(fp2,"%s", a);
}

   if(a[0]=='I')
   {
      //fprintf(fp2,"%s", a);
   strcpy(b,a);
   z1=strlen(b);
fprintf(fp2,"%s",a);

   }

if((a[0]=='E')||(a[0]=='S'))
   {

   strcpy(c,a);
   z2=strlen(c);

   if((z1<z2)||(z1==z2))
      z3=z1;

   if(z1>z2)
      z3=z2;
fprintf(fp2,"%s", c);
                        if(z3>13)
   { p=0;
   for(m=5;m<z3;m++)
         for(m1=0;m1<100000;m1++)
   {    if(m>100)
            break;
         if((b[m1+m-1]=='\n')||(b[m1+m-1]=='\0'))
         break;

                                    for(m2=m1;m2<m1+m;m2++)
{
//fprintf(fp2,"%c", b[m2]);
f[n1]=b[m2];
n1++;
}
// fprintf(fp2,"%s\n", f);
                                       //fprintf(fp2,"\n");
                                       for(m3=0;m3<100000;m3++)
   {
                                          if((c[m3+m-1]=='\n')||(c[m3+m-1]=='\0'))
   break;
                                                   for(m4=m3;m4<m3+m;m4++)
   {
                                                   // fprintf(fp2,"%c", c[m4]);
      h[n2]=c[m4];
      n2++;
   }
                                          //fprintf(fp2,"%s\n%s\n",f,h);
                                                for(n3=0;n3<m;n3++)
{ if((f[n3]!='0')&&(f[n3]!='1')&&(f[n3]!='2')&&(f[n3]!='3')&&(f[n3]!='4')&&(f[n3]!='5')&&(f[n3]!='6')&&(f[n3]!='7')&&(f[n3]!='8')&&(f[n3]!='9'))
{ if(f[n3]==h[n3])
   q=q+5;

                                                         if(f[n3]!=h[n3])
   q=q-4;
}
   }
                                                //  fprintf(fp2,"%d\n",q);
                                                      if(q>p)
{
                                                            p=q;
                                                         strcpy(f1,f);
                                                         strcpy(h1,h);
                                                      k=m1-3;k2=m1+m-4;
                                                      q1=m3-3;p1=m3+m-4;
}
                                          //  fprintf(fp2,"%d\n\n",p);
                                             //fprintf(fp2,"%s\n%s\n",f1,h1);
                                                q=0;

n2=0;

   }

n1=0;

                        for(i=0;i<200;i++)
   {  h[i]='\0';
                           f[i]='\0';

   }

   }
   }
for(i=0;i<5;i++)
{
d[i]='\0';
      d2[i]='\0';
}

for(j=0;j<4;j++)
{
d[j]=b[j];
      d2[j]=c[j];
}
d[4]='\0';
d2[4]='\0';
if(z3>13)
fprintf(fp2,"%s\t%d\t%s\t%d\t%d\n%s\t%d\t%s\t%d\t%d\n%d\t\t%d\n",d,k,f1,k2,z1-4,d2,q1,h1,p1,z2-5,k2-k+1,p);

   for(i=0;i<100000;i++)
   {
   b[i]='\0';
   c[i]='\0';
   }
   z1=0;
   z2=0;
   z3=0;

   }


}
while(a[0]!='@');


for(i=0;i<=8;i++)
e[i]='@';
      e[9]='\0';

fprintf(fp2,"%s\n", e);
      fclose(fp1);
fclose(fp2);


}
3 上面程序输出的最佳局域比对结果
rps-0,B0393.1
I001CAATCAAATTTTTTTTAAGTTAAGTAAAAAGAACACTTAATAAATTATTTTAATATTGTTAATAGTAATGTC
E012CAATCTTAACTTCCAAATGCAGCAATACGTCTACAAGAGACGTTTTGACGGACCAAATATCATCAACGTCAAGAAGACCTGGGAGAAGCTTCTTCTCGCT
I001 3 ATCAAATTTTTTTTAAGTTAAGTAAA 28 73
E012 52 ACCAAATATCATCAACGTCAAGAAGA 77 100
26 40
I002CATTCTGAAAAGATTTATCAACAAAAAAGATTAATAAATTTTAAATC
E034TCAAGGAGCCAAGACTTCTCGTCATCTCTGACCCACGTATCGATCATCAGGCTGTCACTGAGGCTTCCTACGTCGGAGTTCCAGTCATCTCCTTCGTCAA
I002 10 AAGATTTATCAACA 23 48
E034 11 AAGACTTCTCGTCA 24 100
14 34
I003CATATATAAATTAAATATTTTTTACATTTATACATAAACTTAACATC
E056CCGAGTCTCCACTCAAGTTGATCGACATCGGAGTTCCATGCAACAACAAGGGAGAACGCTCAATCGGACTTATGTGGTGGATGCTCGCTCGCGAGATCCT
I003 38 ACTTA 42 48
E056 68 ACTTA 72 100
5 25
rps-2
I001CATCAATAAAATAAGAATTTGGAAGTCTAAAACTTATAAATTTAAAGTC
E012GGACGTGGAGGACGTGGTGGCCGCGGAGGACGTGGTGGACGCGGAGGAAGAGCCGGACGCGGAGGAGAGAAGGAGACCGAATGGACTCCAGTCACCAAGC
I001 21 GGAAG 25 50
E012 46 GGAAG 50 100
5 25
I002CATTCATCAAATAAAAAAATCAAAATAATACCCCAATATTAAAGGTC
E034AACCTTAAAGACGAGGTCCTCAAGATCTCCCCAGTCCAGAAGCAGACCACTGCCGGACAACGTACCCGCTTCAAGGCTTTCGTTGCTATCGGAGATCACG
I002 35 AATATTAAAG 44 48
E034 1 AACCTTAAAG 10 100
10 32