poj 1330 Nearest Common Ancestors

这是一道找最近祖先的问题，我用的是纯暴力的方法。

#include<iostream>
#include<cstdio>
using namespace std;
int father[10024];
int father1[10024];
int father2[10024];
int main()
{
   int n,T,f,s;
   scanf( "%d",&T );
   while( T-- )
   {
     
      scanf( "%d" ,&n ); 
      for( int i=0;i<=n;i++ )
          father[i]=i;
      for( int i=1; i< n; i++ )
      {
           scanf( "%d%d" ,&f,&s );
           father[s]=f;        
      }
      scanf( "%d%d",&f,&s );
      int i=f,j=0,k=0;
      father1[j++]=f;
      while( father[i]!=i )//记录f的所有祖先
      {
         father1[j++]=father[i];
         
//         printf( "father[%d]=%d %d\n",i,father[i],i );
         
         i=father[i];
//         getchar();        
      }
      father1[j++]=father[i];
      i=s;     
      father2[k++]=s;
      while( father[i]!=i )//记录s的所有祖先
      {
         father2[k++]=father[i];
         i=father[i];        
      }
     
      father2[k++]=father[i];
      int flag=0;
      for( int i=0;i<j; i++ )//祖先的匹配
      {
          if( flag ) break;
           for( int t=0;t<k;t++ )
           {
             if( father1[i]==father2[t] )
             {
                   printf( "%d\n",father2[t] );
                   flag=1;
                   break;        
             }    
           }        
      }    
    }
   return 0;    
}

第二种方法：

LCA；这个题好像比上种方法还要慢些；

#include<iostream>
#include<cstdio>
#include<string.h>
#include<string>
#include<vector>
using namespace std;
vector<int> tree[10024],Qes[10024];
int visit[10024],set[10024],indegree[10024],ancestor[10024];
void init( int n )
{
   for( int i=0; i<= n; i++ )
   {
      visit[i]=0;
      set[i]=i;
      indegree[i]=0;
      ancestor[i]=0;
      tree[i].clear();
      Qes[i].clear();    
   }    
}
int find( int x )//查找函数，并压缩路径
{
   return set[x]==x?x:set[x]=find( set[x] );    
}
void Union( int a,int b )//合并函数
{
   int X,Y;
   X=find( a );
   Y=find( b );
   if( X!=Y  )
   {
         set[Y]=X;        
   }      
}
void LCA( int a )
{
   ancestor[a]=a;
   int size=tree[a].size();
   for( int i=0; i< size; i++ )
   {
      LCA( tree[a][i] );
      Union( a,tree[a][i] );
      ancestor[ find( a ) ] = a;   //让a的父节点祖先为a，因为是回溯操作，一定能保证集合的祖先是最近祖先   
    }
    visit[a]=1;
    size=Qes[a].size( );
    for( int i=0;i<size; i++ )
    {//如果已经访问了问题节点,就可以返回结果了.
       if( visit[Qes[a][i]]==1 )
       {//如果这个点处理过，那么这个祖先就是共同祖先  
            cout<<ancestor[find( Qes[a][i] )]<<endl;
            return ;        
       }    
    }    
}
int main( )
{
   int Case,f,s,n;
   scanf( "%d",&Case );
   while( Case-- )
   {
     
      scanf( "%d" ,&n ); 
      init( n );
      for( int i=1; i<n; i++ )
      {
         scanf( "%d %d",&f,&s );
         indegree[s]++;
         tree[f].push_back( s );        
      }
      scanf( "%d %d",&f,&s );

      //相当于询问两次
      Qes[f].push_back( s );
      Qes[s].push_back( f );
      for( int i=1;i<=n ; i++ )
      {//寻找根节点
         if( indegree[i]==0 )
         {
            LCA( i );
            break;        
         }        
       }    
    }
   return 0;    
}