poj1470解题报告

题意：最近公共祖先

题解：最近公共祖先

代码：

1，裸地在线算法

#include <iostream>#include <cstring>#include <cstdio>using namespace std ;const int MAXN = 1000 ;int   N , M ;int   father[MAXN] ;int   count[MAXN]  ;bool  vis[MAXN]    ;int  main(){    int  i , j , a , b ;    int  num , nr , suc ;    char ch ;    while ( scanf( "%d" , & N ) != EOF )    {        memset( vis   , 0 , sizeof( vis   ) ) ;        memset( count , 0 , sizeof( count ) ) ;        for( i = 0 ; i <= N ; i ++ ) father[i] = i ;        for( i = 0 ; i < N ; i ++ )        {            scanf( "%d:(%d)" , & nr , & num ) ;            while( num -- )            {                scanf( "%d" , & suc ) ;                father[suc] = nr ;            }        }        scanf( "%d" , & M ) ;        while( M -- )        {            while( ch = getchar() )            {                if( ch == '(' )                {                     scanf( "%d%d" , & a , & b ) ;                     getchar() ;                     break ;                }            }            while( father[a] != a )            {                vis[a] = 1 ;                a = father[a] ;            }            vis[a] = 1 ;            while( ! vis[b] )                b = father[b] ;            count[b] ++ ;            memset( vis , 0 , sizeof( vis ) ) ;        }        for( i = 0 ; i <= N ; i ++ )            if( count[i] > 0 ) printf( "%d:%d\n" , i , count[i] ) ;    }    return 0 ;}

2，tarjan离线

#include <iostream>#include <cstring>#include <cstdio>using namespace std ;struct type{    int  end , next ;    //int  w ;} ;const int MAXN = 20005 ;int   N , M ;type  edge[MAXN*2] , qedge[MAXN*25] ;int   head[MAXN] , qhead[MAXN] ;int   ind[MAXN]  , vis[MAXN]   ;int   father[MAXN] , ans[MAXN] ;int   Count , qcnt             ;void  addedge( int start , int end ){    edge[Count].end  = end         ;    //edge[Count].w    = w           ;    edge[Count].next = head[start] ;    head[start]      = Count ++    ;}void  qaddedge( int start , int end ){    qedge[qcnt].end  = end          ;    //qedge[qcnt].w    = w            ;    qedge[qcnt].next = qhead[start] ;    qhead[start]     = qcnt ++      ;}int  find( int x ){    if( father[x] == -1 ) return x ;    father[x] = find( father[x] )  ;    return father[x]               ;}void  tarjan( int start ){    int  i , j ;    vis[start] = 1 ;    for( i = qhead[start] ; i != -1 ; i = qedge[i].next )    {        if( vis[qedge[i].end] )        {            int  temp = find( qedge[i].end ) ;            ans[temp] ++ ;        }    }    for( i = head[start] ; i != -1 ; i = edge[i].next )    {        if( !vis[edge[i].end] )        {            tarjan( edge[i].end ) ;            father[edge[i].end] = start ;        }    }}int  main(){    int   i , j ;    int   start , end , num ;    char  ch  ;    while( scanf( "%d" , & N ) != EOF )    {        memset( head   , -1 , sizeof( head   ) ) ;        memset( qhead  , -1 , sizeof( qhead  ) ) ;        memset( father , -1 , sizeof( father ) ) ;        memset( ind    ,  0 , sizeof( ind    ) ) ;        memset( vis    ,  0 , sizeof( vis    ) ) ;        memset( ans    ,  0 , sizeof( ans    ) ) ;        Count = 0 ; qcnt = 0 ;        for( i = 1 ; i <= N ; i ++ )        {            scanf( "%d:(%d)" , & start , & num ) ;            while( num -- )            {                scanf( "%d" , & end ) ;                ind[end] ++ ;                addedge( start , end ) ;                addedge( end , start ) ;            }        }        scanf( "%d" , & M ) ;        while( M -- )        {            while( ch = getchar() )                if( ch == '(' )                {                    scanf( "%d%d" , & start , & end ) ;                    ch = getchar() ;                    break ;                }            qaddedge( start , end ) ;            qaddedge( end , start ) ;        }        for( i = 1 ; i <= N ; i ++ )            if( ind[i] == 0 )            {                tarjan( i ) ;                break ;            }        for( i = 0 ; i <= N ; i ++ )            if( ans[i] ) printf( "%d:%d\n" , i , ans[i] ) ;    }    return 0 ;}