K Best POJ 3111 （二分，最大化平均値）

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 10565 Accepted: 2717Case Time Limit: 2000MS Special Judge

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 21 11 21 3

Sample Output

1 2

Source

分析：稍后

代码如下：

/* Author:kzl */#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>#include<cmath>using namespace std;const int maxx = 100000;const double EPS = 1e-6;int n,k;struct My{int v,w;}cun[maxx];struct MM{double val;int id;}aa[maxx];bool cmp(MM a,MM b){return a.val>b.val;}bool flag(double num,bool fl){    double sum = 0;for(int i=0;i<n;i++){    aa[i].val = cun[i].v - num * cun[i].w;    aa[i].id = i+1;}sort(aa,aa+n,cmp);if(fl)printf("%d",aa[0].id);for(int i=0;i<k;i++){    sum+=aa[i].val;    if(fl&&i)printf(" %d",aa[i].id);}if(fl)printf("\n");if(sum>=0)return true;else return false;}int main(){while(scanf("%d%d",&n,&k)!=EOF){    for(int i=0;i<n;i++)    {        scanf("%d%d",&cun[i].v,&cun[i].w);    }    double l = 0,r = 1e7,mid = 0;    while(fabs(r-l)>EPS){        mid = (l+r)/2.0;        if(flag(mid,false))l = mid;        else r = mid;    }    flag(mid,true);}return 0;}