K Best（最大化平均数）_二分搜索

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

【题意】有n个物品的重量和价值分别是w[i]和v[i]，从中选出K个物品使得单位重量的价值最大。

1<=k<=n<=10^5

1<=w[i],v[i]<=10^7

【思路】一般想到的是按单位价值对物品排序，然后贪心选取，但是这个方法是错误的，对于有样例不满足。

我们一般用二分搜索来做（其实这就是一个01分数规划）

我们定义：

条件 C(x) =可以选k个物品使得单位重量的价值不小于x。

因此原问题转换成了求解满足条件C(x)的最大x。那么怎么判断C(x)是否满足？

变形：（SUM(v[i])/SUM(w[i])）>=x (i 属于我们选择的某个物品集合S)

进一步：SUM(v[i]-x*w[i])>=0

于是：条件满足等价于选最大的k个和不小于0.于是排序贪心选择可以判断，每次判断的复杂度是O(nlogn)。

参考链接：http://www.2cto.com/kf/201308/235786.html

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int inf=0x3f3f3f3f;
const int N=100010;
struct node
{
    double w,v,ret;
    int id;
}a[N];
int n,k;
bool cmp(node a,node b)
{
    return a.ret>b.ret;
}
bool ok(double mid)//是否满足条件SUM(v[i]-x*w[i])>=0
{
    double sum=0;
    for(int i=1;i<=n;i++)
    {
        a[i].ret=a[i].v-mid*a[i].w;
    }
    sort(a+1,a+1+n,cmp);
    for(int i=1;i<=k;i++)
    {
        sum+=a[i].ret;
    }
    return sum>=0;
}
int main()
{
   while(~scanf("%d%d",&n,&k))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&a[i].v,&a[i].w);
            a[i].id=i;
        }
         double l=0,r=1.0*inf;
        while(r-l>1e-5)//二分搜索
        {
            double mid=(l+r)/2;
            if(ok(mid))
                l=mid;
            else r=mid;
        }
        for(int i=1;i<k;i++)
        {
            printf("%d ",a[i].id);
        }
        printf("%d\n",a[k].id);

    }
    return 0;
}