[POJ 2976]Dropping tests（0-1分数规划）

最新推荐文章于 2019-03-06 12:53:30 发布

转载最新推荐文章于 2019-03-06 12:53:30 发布 · 55 阅读

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原文链接：http://www.cnblogs.com/Zars19/p/6942372.html

本文介绍了一种使用0-1分数规划算法来解决如何通过丢弃特定数量的测试成绩以提高累计平均分的问题。该算法首先设定能达到的最大值p，并通过二分查找法确定最优解。文中提供了一个具体的示例，展示了当允许丢弃某些测试成绩时，如何显著提高累计平均分。

Description

In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be.

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

Solution

0-1分数规划裸题

设能够达到的最大值为p

则有∑(ai*xi)/∑(bi*xi)<=p

∑ai*xi-∑bi*xi*p<=0

即∑xi(ai-bi*p)<=0

也就是说∑xi(ai-bi*p)的最大值为0

二分答案

如果p<mid ∑xi*(ai-bi*mid)的最大值<0

如果p>mid ∑xi*(ai-bi*mid)的最大值>0

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define eps 1e-10
using namespace std;
int n,k;
double a[1005],b[1005],c[1005];
int main()
{
    while(~scanf("%d %d",&n,&k)&&n)
    {
        for(int i=1;i<=n;i++)scanf("%lf",&a[i]);
        for(int i=1;i<=n;i++)scanf("%lf",&b[i]);
        double l=0,r=1,mid;
        while(r-l>eps)
        {
            mid=(l+r)/2;
            for(int i=1;i<=n;i++)c[i]=a[i]-b[i]*mid;
            sort(c+1,c+1+n);
            double res=0;
            for(int i=k+1;i<=n;i++)
            res+=c[i];
            if(res>0)l=mid;else r=mid;
        }
        printf("%.0f\n",mid*100);
        
    }
    return 0;
}