如何在双向图中找到两条不相交的路径?

我有一个包含 14 个节点和 21 个链接的图表。此图显示了一个光纤网络。这些链接是双向的，每个链接上都有一些资源。假设有一个从源到目标的工作路径，该路径携带包含数据的数据包并使用一些资源(它所遍历的链路的一定带宽(。对于每个源和目标，我必须提前同时建立工作和保护路径，以防链路故障，但它们是链路脱节的。(它们不能遍历任何公共链接(

例如，我通过 route<1,2,3,4> 将数据包从节点 1 发送到节点 4 作为工作路径。 如果链路 1-2 出现故障，我必须通过预先建立的保护路径发送数据包，并且与工作路径脱节。例如，我的保护路径可能是 <1,9,3,4>。

目的是用 C/C++ 编写代码来查找与工作路径不相交的链接的保护路径。实际上，我无法得到这样做的想法。任何帮助将不胜感激。

这是我的代码片段，我将资源分配给工作路径，我不知道如何在必须与工作路径脱节的情况下为建立保护路径做同样的事情。

int required_fs;
int des;
int idpk;
double holding;
int src;
//get the information about the packet that is sent from source to destination.    
Packet *pkptr;
pkptr = op_pk_get (0);
op_pk_nfd_get(pkptr,"bw_fs",&required_fs);
op_pk_nfd_get(pkptr,"des",&des);
op_pk_nfd_get(pkptr,"src",&src);
op_pk_nfd_get(pkptr,"id",&idpk);
op_pk_nfd_get(pkptr,"ht",&holding);

bw_req=bw_req + required_fs;
number_of_req=number_of_req+1;
if (number_of_req > 1000000)
{
FILE* file1=fopen("C:/400f_rsa.txt","a+");
fprintf(file1,"n");
fprintf(file1,"number_of_req ,number_of_nack,number_of_ack , bw_req , bw_nack , bw_ack " );
fprintf(file1,"n");
fprintf(file1,"  %d , %d , %d , %d , %d , %d   ", number_of_req, number_of_nack ,number_of_ack,bw_req,bw_nack,bw_ack );
fprintf(file1,"n");
fprintf(file1,"------------------------------- ");
fclose (file1);
op_sim_end("1000000 paket","","","");
}
//  Allocate the resources on each link to the working path, This part must be the same for the protection path too.
int determined_t=0;
int determined_r=0;
int determined_p_f=0;
int determined_p_e=0;
int determined_k=0;
for ( int i=1; i<=10; i++)
{
if (transmitter[src][i]==0)
{
determined_t=i;
break;
}
}
if (determined_t!=0)
{
for ( int i=1; i<=10; i++)
{
if (reciever[des][i]==0)
{
determined_r=i;
break;
}
}
if (determined_r!=0)
{
for ( int k=1; k<=2 ; k++)
{
determined_p_f=0;
determined_p_e=0;
int count = paths_node[src][des][k][2][14];
zero_array();
for (int i=1; i<=count; i++)
{
finding_fs( k, i, des, src );
}
if ( ff_rr==0)
{//ff
////checking gap
determined_p_f=find_first_free_gap(required_fs);
if (determined_p_f!=0)
{   
determined_p_e=determined_p_f+required_fs-1;
if (determined_p_e != 1000)
{
determined_p_e=determined_p_e+1;
}
determined_k=k;
break;
}

}
else if (ff_rr==1)
{//rr
clear_rr_gap();
find_rr_gap(required_fs);
int index=rr_spectrum();
determined_p_f=ary_rr[index].first;
if (determined_p_f!=0)
{
determined_p_e=ary_rr[index].last;
determined_k=k;
break;
}
}

}
if (determined_p_f!=0)
{
//add to ls , applying
int count_link = paths_node[src][des][determined_k][2][14];
for ( int i=1; i<=count_link ; i++)
{
int num_link_p=paths_node[src][des][determined_k][2][i];
for ( int j=determined_p_f ; j<=determined_p_e; j++)
{
links_fs[num_link_p][j]=1;
}
}
reciever[des][determined_r]=1;
transmitter[src][determined_t]=1;
ls(determined_p_f,determined_p_e,determined_r,determined_t,determined_k,src,des,idpk);
number_of_ack= number_of_ack +1 ;
bw_ack= bw_ack + required_fs;
op_intrpt_schedule_self(op_sim_time ()+ holding,idpk);
op_pk_destroy(pkptr);
}
else
{
number_of_nack=number_of_nack+1;
bw_nack= bw_nack + required_fs;
op_pk_destroy(pkptr);
}
}
else
{
number_of_nack=number_of_nack+1;
bw_nack= bw_nack + required_fs;
op_pk_destroy(pkptr);
}
}
else
{
number_of_nack=number_of_nack+1;
bw_nack= bw_nack + required_fs;
op_pk_destroy(pkptr);
}