POJ3111 K Best

Description

Demy has n jewels. Each of her jewels has some value v_i and weight w_i.

Since her husband John got broke after recent financial crises, Demy has decided to sell some jewels. She has decided that she would keep k best jewels for herself. She decided to keep such jewels that their specific value is as large as possible. That is, denote the specific value of some set of jewels S = {i₁, i₂, …, i_k} as

.

Demy would like to select such k jewels that their specific value is maximal possible. Help her to do so.

Input

The first line of the input file contains n — the number of jewels Demy got, and k — the number of jewels she would like to keep (1 ≤ k ≤ n ≤ 100 000).

The following n lines contain two integer numbers each — v_i and w_i (0 ≤ v_i ≤ 10⁶, 1 ≤ w_i ≤ 10⁶, both the sum of all v_i and the sum of all w_i do not exceed 10⁷).

Output

Output k numbers — the numbers of jewels Demy must keep. If there are several solutions, output any one.

Sample Input

3 2
1 1
1 2
1 3

Sample Output

1 2
最近刚学习了01规划，嗯，这差不多是裸题，然后，没有用binary_search ,学习了Dinkelbach ，其实感觉很迷吧， 会用就行了？
嗯，后面会多练习几道类似的题 。

#include <algorithm>
#include <iostream>
#include <cstdio>
const int N = 100000 + 11 ; 
using namespace std ;
int n , k , maxn ; 
struct node
{
    int v , w , id ; double val ;
    bool operator < ( const node a ) const { return val > a.val ; }
} num[N] ;

void Init( )
{
    scanf( "%d%d" , &n , &k ) ;
    for( int i = 1 ; i <= n ; ++i )
    {    scanf( "%d%d" , &num[i].v , &num[i].w ) ;    
        num[i].id = i ;
    }
}

double abse( double a )
{
    return ( a < 0 ) ? -a : a ;
}

double search( double l )
{
    for( int x = 1 ; x <= n ; ++x ) num[x].val = (double)num[x].v - (double)num[x].w * l ;
    sort( num + 1 , num + 1 + n ) ;
    int tv = 0 , tw = 0 ;
    for( int i = 1 ; i <= k ; ++i )
        tv += num[i].v , tw += num[i].w ;    
    double ret = (double)tv/(double)tw ;
    return ret ; 
}


void Solve( )
{
    double ans = 1 , tmp ; 
    while( 1 )
    {
        tmp = search( ans ) ;
        if( abse( ans - tmp ) < 0.001 ) break ;
        ans = tmp ;
    }
    for( int x = 1 ; x <= k ; ++x )
        printf( "%d " , num[x].id ) ;
    puts( "" ) ;
    
}

int main( )
{
    Init( ) ;
    Solve( ) ;
    return 0 ;
}