二分三分 D POJ-3111 K Best

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

n 个珍珠选k个使其s价值最大，如果每个都算s排序取前k个的话是不符合题意的，因为这无法保证总s价值最大，一开始想了好久没想过来，看了几篇题解才懂了。通过x（s）=v/w可得，y=x*w-v=0时，可得到最大s价值，所以我们可以二分x，计算每个的y并排序，即可得到答案。

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

const double inf=99999999.0;
struct jew{
    long v;
    long w;
    int num;
    double y;
}j[100000];
int k,n;
double s;
const double eps=1e-8;

double cal(double x);
bool cmp(jew j1,jew j2);

int main()
{
    int i;
    double left,right,mid;

    scanf("%d%d",&n,&k);

    for(i=0;i<n;i++)
    {
        scanf("%d%d",&j[i].v,&j[i].w);
        j[i].num=i+1;
    }
    left=0;
    right=inf;    //注：左右取值
    while(left+eps<right){
        mid=(left+right)/2;
        if(cal(mid)>0)
            left=mid;
        else
            right=mid;
    }
    for(i=0;i<k-1;i++)
        printf("%d ",j[i].num);
    printf("%d\n",j[i].num);
}

double cal(double x)    二分x计算y
{
    int i;

    for(i=0;i<n;i++)
        j[i].y=j[i].v-x*j[i].w;
    sort(j,j+n,cmp);
    s=0;//!
    for(i=0;i<k;i++)
        s+=j[i].y;
    return s;
}

bool cmp(jew j1,jew j2)    //结构体排序比较函数
{
    return j1.y>j2.y;
}