BZOJ4476 送礼物

这道题真是有趣呀。

其实就是一个分数规划问题，用一个二分加log来得去掉分母。

分四种情况讨论

1.lenth > L && num ( max ) > num ( min )

2.lenth > L && num ( max ) < num ( min )

3.lenth == L && num ( max ) > num ( min )

4.lenth == L && num ( max ) < num ( min )

1,2种情况中最大值与最小值一定在选择序列两端，设左右端点编号为i,j

1. ( Vi - Vj ) / ( i - j + k ) = ans -> ( Vi - i * ans ) - ( Vj - j * ans ) = k * ans

二分ans并check左式最大值是否大于右式

2.( Vj - Vi ) / ( i - j + k ) = ans -> ( Vj + j * ans ) - ( Vi + i * ans ) = k * ans

同情况1

3,4用单调队列for一遍就可以了

#include<cstdio>
#include<algorithm>
#include<cassert>
//#define DEBUG
using namespace std ; 

const int MAXN = 5 * 100000 + 20 ;
int N , K , L , R ; 
long long a [ MAXN ] ;
int p [ MAXN ] ;

template < class T1 , class T2 > 
void max_equal ( T1 & a , const T2 & b ) {
    if ( a < b ) a = b ; 
}

double solve1 ( const double M ) {
#define f(i) (a[i]-M*(i))
    double ans = -1e9 ;
    int l = 0 , r = -1 ; 
    for ( int i = L ; i <= N ; ++ i ) {
        while ( r - l + 1 >= 1 && i - p [ l ] + 1 > R ) ++ l ;  
        while ( r - l + 1 >= 1 && ! ( f ( p [ r ] ) < f ( i - L + 1 ) ) ) -- r ;
        p [ ++ r ] = i - L + 1 ;
        max_equal ( ans , f ( i ) - f ( p [ l ] ) ) ;
    } 
    return ans ;
#undef f 
}

double Solve1 () {
    double L = 0 , R = 1e4 ; 
    while ( R - L >= 1e-10 ) {
        const double M = ( L + R ) / 2 ; 
        if ( solve1 ( M ) >= K * M ) L = M ; 
        else R = M ; 
    }
    return ( L + R ) / 2 ; 
}

double solve2 ( const double M ) {
#define f(i) (a[i]+M*(i))
    double ans = -1e9 ; 
    int l = 0 , r = -1 ; 
    for ( int i = L ; i <= N ; ++ i ) {
        while ( r - l + 1 >= 1 && i - p [ l ] + 1 > R ) ++ l ;
        while ( r - l + 1 >= 1 && ! ( f ( p [ r ] ) > f ( i - L + 1 ) ) ) -- r ; 
        p [ ++ r ] = i - L + 1 ; 
        max_equal ( ans , f ( p [ l ] ) - f ( i ) ) ; 
    }
    return ans ;
#undef f
}

double Solve2 () {
    double L = 0 , R = 1e4 ; 
    while ( R - L >= 1e-10 ) {
        const double M = ( L + R ) / 2 ; 
        if ( solve2 ( M ) >= K * M ) L = M ; 
        else R = M ; 
    }
    return ( L + R ) / 2 ; 
}

double Solve3 () {
    double ans = -1e9; 
    int l = 0 , r = -1 ; 
    for ( int i = 1 ; i <= N ; ++ i ) {
        while ( r - l + 1 >= 1 && i - p [ l ] + 1 > L ) ++ l ;
        while ( r - l + 1 >= 1 && ! ( a [ p [ r ] ] < a [ i ] ) ) -- r ; 
        p [ ++ r ] = i ; 
        max_equal ( ans , a [ i ] - a [ p [ l ] ] ) ; 
    }
    return ans / ( L - 1 + K ) ;
}

double Solve4 () {
    double ans = -1e9; 
    int l = 0 , r = -1 ; 
    for ( int i = 1 ; i <= N ; ++ i ) {
        while ( r - l + 1 >= 1 && i - p [ l ] + 1 > L ) ++ l ;
        while ( r - l + 1 >= 1 && ! ( a [ p [ r ] ] > a [ i ] ) ) -- r ; 
        p [ ++ r ] = i ; 
        max_equal ( ans , a [ p [ l ] ] - a [ i ] ) ; 
    }
    return ans / ( L - 1 + K ) ;
}

double solve () {
    scanf ( "%d%d%d%d" , & N , & K , & L , & R ) ; 
    for ( int i = 1 ; i <= N ; ++ i ) scanf ( "%lld" , & a [ i ] ) ;
    return max ( max ( Solve1 () , Solve2 () ) , max ( Solve3 () , Solve4 () ) ) ;
}

int main () {
    int T ; 
    scanf ( "%d" , & T ) ; 
    while ( T -- ) printf ( "%.4lf
" , solve () ) ; 
    return 0 ; 
}