BZOJ-2726: [SDOI2012]任务安排(DP+平衡树维护凸壳)

题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2726

动态转移方程:f(j)=min{f(j)+【sumt(i)-sumt(j)+S】*sumf(i)}

sum(i)表示从i到n的和。

然后这个方程就可以用一个平衡树来维护一个决策的下凸壳,然后就做到O(n log n),然后就可以A了。

代码(可怜我的treap居然比set还慢 555):

#include 

#include 

#include 

#include 

 

using namespace std ;

 

#define rep( i , x ) for ( int i = 0 ; i ++ < x ; )

#define down( i , x ) for ( int i = x ; i ; -- i )

 

#define L( t ) left[ t ]

#define R( t ) right[ t ]

#define pre( t ) prefix[ t ]

#define suff( t ) suffix[ t ]

#define P( t ) priority[ t ]

#define K( t ) key[ t ]

 

const int maxn = 300100 ;

 

typedef long long ll ;

typedef long double ld ;

 

ll inf = 1000000000 ;

ll INF = inf * inf ;

 

struct point {

     

    ld x , y ;

    int pos ;

     

    void oper( ld _x , ld _y , int _pos ) {

        x = _x , y = _y , pos = _pos ;

    }

     

    bool operator < ( const point &a ) const {

        return x < a.x ;

    }

     

    bool operator == ( const point &a ) const {

        return x == a.x ;

    }

     

    bool operator > ( const point &a ) const {

        return x > a.x ;

    }

     

    ld cal( point rec ) {

        return ( y - rec.y ) / ( x - rec.x ) ;

    }

     

} key[ maxn ] ;

 

int left[ maxn ] , right[ maxn ] , prefix[ maxn ] , suffix[ maxn ] , priority[ maxn ] ;

int V , roof ;

 

void Left( int &t ) {

    int k = R( t ) ;

    R( t ) = L( k ) ;

    L( k ) = t ;

    t = k ;

}

 

void Right( int &t ) {

    int k = L( t ) ;

    L( t ) = R( k ) ;

    R( k ) = t ;

    t = k ;

}

 

void Insert( point k , int &t ) {

    if ( ! t ) {

        t = ++ V ;

        K( t ) = k , L( t ) = R( t ) = 0 , P( t ) = rand(  ) ;

        return ;

    }

    if ( k < K( t ) ) {

        Insert( k , L( t ) ) ;

        if ( P( L( t ) ) > P( t ) ) Right( t ) ;

    } else {

        Insert( k , R( t ) ) ;

        if ( P( R( t ) ) > P( t ) ) Left( t ) ;

    }

}

 

void Delete( point k , int &t ) {

    if ( K( t ) == k ) {

        if ( ! L( t ) ) {

            t = R( t ) ; return ;

        } else if ( ! R( t ) ) {

            t = L( t ) ; return ;

        } else {

            if ( P( L( t ) ) > P( R( t ) ) ) {

                Right( t ) ; Delete( k , R( t ) ) ;

            } else {

                Left( t ) ; Delete( k , L( t ) ) ;

            }

        }

    } else Delete( k , k < K( t ) ? L( t ) : R( t ) ) ;

}

 

int Prefix( point k ) {

    int temp = 0 ;

    for ( int t = roof ; t ; t = k < K( t ) ? L( t ) : R( t ) ) {

        if ( k > K( t ) && ( ! temp || K( t ) > K( temp ) ) ) temp = t ;

    }

    return temp ;

}

 

int Suffix( point k ) {

    int temp = 0 ;

    for ( int t = roof ; t ; t = k < K( t ) ? L( t ) : R( t ) ) {

        if ( k < K( t ) && ( ! temp || K( t ) < K( temp ) ) ) temp = t ;

    }

    return temp ;

}

 

int Find( point k ) {

    for ( int t = roof ; t ; t = k < K( t ) ? L( t ) : R( t ) ) {

        if ( K( t ) == k ) return t ;

    }

    return 0 ;

}

 

void Push( point k ) {

    int rec = Find( k ) ;

    if ( rec ) {

        if ( K( rec ).y <= k.y ) return ;

        Delete( K( rec ) , roof ) ;

    }

    int p = Prefix( k ) , s = Suffix( k ) ;

    if ( k.cal( K( p ) ) >= K( p ).cal( K( s ) ) ) return ;

    int ret ;

    for ( ; K( p ).x > - inf ; ) {

        ret = pre( p ) ;

        if ( k.cal( K( ret ) ) <= K( ret ).cal( K( p ) ) ) {

            Delete( K( p ) , roof ) ;

            p = ret ;

        } else break ;

    }

    for ( ; K( s ).x < inf ; ) {

        ret = suff( s ) ;

        if ( k.cal( K( s ) ) >= k.cal( K( ret ) ) ) {

            Delete( K( s ) , roof ) ;

            s = ret ;

        } else break ;

    }

    Insert( k , roof ) ;

    suff( p ) = V , pre( s ) = V , pre( V ) = p , suff( V ) = s ;

}

 

point Search( ld k , int t ) {

    if ( K( t ).x == - inf ) return Search( k , R( t ) ) ;

    if ( K( t ).x == inf ) return Search( k , L( t ) ) ;

    ld lk = K( t ).cal( K( pre( t ) ) ) , rk = K( t ).cal( K( suff( t ) ) ) ;

    if ( k >= lk && k <= rk ) return K( t ) ;

    if ( k < lk ) return Search( k , L( t ) ) ;

    if ( k > rk ) return Search( k , R( t ) ) ;

}

 

ll f[ maxn ] , n , S , T[ maxn ] , F[ maxn ] , sumt[ maxn ] , sumf[ maxn ] ;

 

void Init_treap(  ) {

    V = 2 , roof = 1 ;

    P( 1 ) = rand(  ) , P( 2 ) = rand(  ) ;

    K( 1 ).oper( ld( - inf ) , ld( INF ) , - 1 ) , K( 2 ).oper( ld( inf ) , ld( INF ) , - 1 ) ;

    pre( 1 ) = suff( 2 ) = 0 , suff( 1 ) = 2 , pre( 2 ) = 1 ;

    L( 1 ) = R( 1 ) = L( 2 ) = R( 2 ) = 0 ;

    if ( P( 1 ) > P( 2 ) ) {

        R( 1 ) = 2 ;

        roof = 1 ;

    } else {

        L( 2 ) = 1 ;

        roof = 2 ;

    }

}

 

point make( int x ) {

    point temp ;

    temp.oper( ld( sumt[ x ] ) , ld( f[ x ] ) , x ) ;

    return temp ;

}

 

int main(  ) {

    srand( 1221 ) ;

    scanf( "%lld%lld" , &n , &S ) ;

    rep( i , n ) scanf( "%lld%lld" , T + i , F + i ) ;

    sumt[ n + 1 ] = sumf[ n + 1 ] = 0 ;

    down( i , n ) {

        sumt[ i ] = sumt[ i + 1 ] + T[ i ] ;

        sumf[ i ] = sumf[ i + 1 ] + F[ i ] ;

    }

    Init_treap(  ) ;

    f[ n + 1 ] = 0 ;

    Push( make( n + 1 ) ) ;

    down( i , n ) {

        point rec = Search( ld( sumf[ i ] ) , roof ) ;

        f[ i ] = f[ rec.pos ] + ( sumt[ i ] - sumt[ rec.pos ] + S ) * sumf[ i ] ;

        Push( make( i ) ) ;

    }

    printf( "%lld\n" , f[ 1 ] ) ;

    return 0 ;

}


你可能感兴趣的:(BZOJ-2726: [SDOI2012]任务安排(DP+平衡树维护凸壳))