(1)分别对graph A和graph B用DFS分割出各个"孤立的group",统计出每个group的元素数目(number) 
(2)for each group ga in graph A do 
  {   
        for each element ea in ga do
        {
           record each track's length by 4 directions in a table ta 
        }
        find = false;
        for each group gb which element number is equal to ga in graph B do
        {
                 for each element eb in gb do 
                 {
                     record each track's length by 4 directions in table tb
                 }
        if(ta==tb) 
          {
           find=true;
           break;
           } 
        } 
  if (find==false) cout<<"NO"<<endl;
}
   cout<<"YES"<<endl;

• 如此弱的数据，是算法就能过，不用证明 (13) fanhqme 2008-09-29 17:36:17