美文网首页信息学竞赛题解(IO题解)
BZOJ-2002: [Hnoi2010]Bounce 弹飞绵羊

BZOJ-2002: [Hnoi2010]Bounce 弹飞绵羊

作者: AmadeusChan | 来源:发表于2018-10-16 20:19 被阅读0次

    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2002

    每个点只会跳到另外一个确切的点,这跟树中每个节点只有一个父节点一样的,那么按照这个规律建树即可,然后LCT维护,对于每个查询,ACCESS(V),然后输出sizeleft[v]即可。

    代码:

    472309f790529822e608a4b3d5ca7bcb0a46d483.jpg.png
    # include <cstdio>
    # include <algorithm>
    # include <cstring>
     
    using namespace std ;
     
    # define MAXN 200010
    # define L( t ) left[ t ]
    # define R( t ) right[ t ]
    # define F( t ) father[ t ]
    # define G( t )F(F( t ) )
    # define S( t ) size[ t ]
    # define P( t ) parent[path_roof( t ) ]
    # define update( t )S( t ) =S(L( t ) ) +S(R( t ) ) + 1
    # define C( t ) (L(F( t ) ) == t )
    # define clear( t )memset( t , 0 , sizeof( t ) )
     
    int left[ MAXN ] , right[ MAXN ] , father[ MAXN ] , parent[ MAXN ] , size[ MAXN ] ;
     
    void zag( int t ) {
        int k =R( t ) , u =F( t ) ;
        bool flag =C( t ) ;
        R( t ) =L( k ) ;F(L( k ) ) = t ;update( t ) ;
        L( k ) = t ;F( t ) = k ;           update( k ) ;
        F( k ) = u ; if ( u ) if ( flag )L( u ) = k ; else R( u ) = k ;
    }
     
    void zig( int t ) {
        int k =L( t ) , u =F( t ) ; 
        bool flag =C( t ) ;
        L( t ) =R( k ) ;F(R( k ) ) = t;update( t ) ;
        R( k ) = t ;F( t ) = k ;          update( k ) ;
        F( k ) = u ; if ( u ) if ( flag )L( u ) = k ; else R( u ) = k ;
    }
     
    void splay( int t ) {
        while (F( t ) ) {
            if ( !G( t ) ) if (C( t ) )zig(F( t ) ) ; else zag(F( t ) ) 
            ; else {
                if (C( t ) ) {
                    if (C(F( t ) ) )zig(G( t ) ) ;
                    zig(F( t ) ) ;
                } else {
                    if ( !C(F( t ) ) )zag(G( t ) ) ;
                    zag(F( t ) ) ;
                }
            }
        }
    }
     
    int path_roof( int t ) {
        splay( t ) ;
        while (L( t ) ) t =L( t ) ;
        return t ;
    }
     
    void Access( int t ) {
        int v = 0 ;
        do {
            splay( t ) ;
            F(R( t ) ) = 0 ;P(R( t ) ) = t ;R( t ) = v ;F( v ) = t ;update( t ) ;
            v = t ; t =P( t ) ;
        } while ( t ) ;
    }
     
    void Make_tree( int t ) {
        L( t ) =R( t ) =F( t ) = parent[ t ] = 0 ;
        S( t ) = 1 ;
    }
     
    void Cut( int t ) {
        Access( t ) ;splay( t ) ;
        F(L( t ) ) = 0 ;L( t ) = 0 ;P( t ) = 0 ;update( t ) ;
    }
     
    void Join( int t , int v ) {
        P( t ) = v ; 
        Access( t ) ;
    }
     
    void Init(  ) {
        F( 0 ) =L( 0 ) =R( 0 ) =S( 0 ) = 0;
        clear( parent ) ;
    }
     
    int k[ MAXN ] , n , m ;
     
    int main (  ) {
        Init(  ) ; 
        scanf( "%d" , &n ) ;
        for ( int i = 0 ; i ++ < n + 1 ; )Make_tree( i ) ;
        for ( int i = 0 ; i ++ < n ; ) {
            scanf( "%d" , &k[ i ] ) ;
            Join( i ,min( n + 1 , i + k[ i ] ) ) ;
        }
        scanf( "%d" , &m ) ;
        while ( m -- ) {
            int x ;
            scanf( "%d" , &x ) ;
            if ( x == 1 ) {
                int y ;
                scanf( "%d" , &y ) ; y ++ ;
                Access( y ) ;splay( y ) ;
                printf( "%d\n" ,S(L( y ) ) ) ;
            } else {
                int y , z ;
                scanf( "%d%d" , &y , &z ) ; y ++ ;
                Cut( y ) ;Join( y ,min( y + z , n + 1 ) ) ;
            }
        }
        return 0 ;
    }
    

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