（POJ - 3111）K Best

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本文解析了POJ-3111 KBest问题，介绍了一种通过二分查找来最大化平均价值比的方法。面对一定数量的珠宝，需选择k个使得总价值与总重量比值最大，采用二分法逼近最优解。

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（POJ - 3111）K Best

Time Limit: 8000MS Memory Limit: 65536K

Total Submissions: 11274 Accepted: 2901

Case Time Limit: 2000MS Special Judge

Description

Demy has n jewels. Each of her jewels has some value vi and weight wi.

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, …, ik} as
这里写图片描述
.

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 — vi and wi (0 ≤ vi ≤ 106, 1 ≤ wi ≤ 106, both the sum of all vi and the sum of all wi 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

题目大意：有n个珠宝，现在要保留k个，使得 $sigma(v[i])/sigma(w[i])$ 最大。

思路：二分，最大化平均值。构造函数 $sigma(y[i])=sigma(v[i])-x*sigma(w[i])$ ，二分x的值，二分的终点便是使 $sigma(y[i])$ 接近0的x值，x的上界是max{v[i]/w[i]}。

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

const double eps=1e-8;
const int maxn=100005;
int n,k;

struct node
{
    int id;
    double v,w,y;
}a[maxn];

bool cmp(node a,node b)
{
    return a.y>b.y;
}

bool check(double x)
{
    for(int i=0;i<n;i++) a[i].y=a[i].v-x*a[i].w;
    sort(a,a+n,cmp);
    double sum=0.0;
    for(int i=0;i<k;i++) sum+=a[i].y;
    return sum>=0;
}

int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        double lo=0.0,hi=0.0,mid;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&a[i].v,&a[i].w);
            hi=max(hi,a[i].v/a[i].w);
            a[i].id=i+1;
        }
        while(hi-lo>eps)
        {
            mid=(hi+lo)/2.0;
            if(check(mid)) lo=mid;
            else hi=mid;
        }
        for(int i=0;i<k;i++)
        {
            if(i==k-1) printf("%d\n",a[i].id);
            else printf("%d ",a[i].id);
        }
    }
    return 0;
}