【POJ】2976 Dropping tests 01分数规划

传送门：【POJ】2976 Dropping tests

题目大意：给你长度为n的一对整数a[],b[]（注意是一对的），根据式子可以得到：∑a[ i ] / ∑b[ i ]，现在给你整数k，你可以从n个中剔除k对，问剩下的根据式子能得到的最大值是多少，答案*100并且四舍五入精确到个位。

题目分析：

很清晰的01分数规划，设Q(L) = ∑a[ i ] - L * ∑b[ i ]。则Q(L) < 0时能得到更优解，Q(L) > 0时不能得到最优解，Q(L) = 0时是解。

用二分就Q(L) < 0的时候修改下界，Q(L) > 0的时候修改上界。注意精度问题即可。

二分代码如下：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clear( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1005 ;
const double INF = 1e18 ;
const double eps = 1e-6 ;

int a[MAXN] , b[MAXN] ;
double c[MAXN] ;
int n , k ;

double solve ( double r ) {
	REP ( i , n )
		c[i] = a[i] - r * b[i] ;
	sort ( c , c + n ) ;
	double ans = 0 ;
	REPF ( i , k , n - 1 )
		ans += c[i] ;
	return ans ;
}

void work () {
	while ( ~scanf ( "%d%d" , &n , &k ) && ( n || k ) ) {
		REP ( i , n )
			scanf ( "%d" , &a[i] ) ;
		REP ( i , n )
			scanf ( "%d" , &b[i] ) ;
		double l = 0 , r = INF , m ;
		while ( fabs ( r - l ) > eps ) {
			double m = ( l + r ) / 2 ;
			if ( solve ( m ) >= eps )
				l = m ;
			else
				r = m ;
		}
		printf ( "%d\n" , ( int ) ( 100 * l + 0.5 ) ) ;
	}
}

int main () {
	work () ;
	return 0 ;
}

然后是迭代的代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clear( a , x ) memset ( a , x , sizeof a )

const int MAXN = 1005 ;
const double INF = 1e18 ;
const double eps = 1e-5 ;

struct Node {
	int a , b ;
	double c ;
	bool operator < ( const Node &t ) const {
		return c > t.c ;
	}
} ;

Node node[MAXN] ;
int n , k ;

double solve ( double r ) {
	REP ( i , n )
		node[i].c = node[i].a - r * node[i].b ;
	sort ( node , node + n ) ;
	double A = 0 , B = 0 ;
	REP ( i , n - k )
		A += node[i].a , B += node[i].b ;
	return A / B ;
}

void work () {
	while ( ~scanf ( "%d%d" , &n , &k ) && ( n || k ) ) {
		REP ( i , n )
			scanf ( "%d" , &node[i].a ) ;
		REP ( i , n )
			scanf ( "%d" , &node[i].b ) ;
		double res = 0 , tmp ;
		while ( 1 ) {
			tmp = solve ( res ) ;
			if ( fabs ( res - tmp ) <= eps )
				break ;
			res = tmp ;
		}
		printf ( "%.0f\n" , 100 * res ) ;
	}
}

int main () {
	work () ;
	return 0 ;
}