POJ3111 K Best

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 10261 Accepted: 2644Case 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

Northeastern Europe 2005, Northern Subregion

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题目的的意思是给出n个分数取其中m个，要求分子之和与分母之和之比最大

思路：二分+验证

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <string>#include <vector>using namespace std;#define inf 0x3f3f3f3f#define LL long longint n,m;struct node{    int a,b,id;} p[100005];double k;bool cmp(node a,node b){    double x=a.a-(a.b*k);    double y=b.a-(b.b*k);    return x>y;}bool ok(double mid){    k=mid;    sort(p,p+n,cmp);    LL x=0,y=0;    for(int i=0; i<m; i++)    {        x+=p[i].a;        y+=p[i].b;    }    double ans=x*1.0/y;    if(ans>=mid)        return 1;    return 0;}int main(){    while(~scanf("%d%d",&n,&m)&&(m||n))    {        for(int i=0; i<n; i++)            scanf("%d%d",&p[i].a,&p[i].b),p[i].id=i+1;        double l=0,r=1e6;        while(r-l>1e-5)        {            double mid=(l+r)/2;            if(ok(mid))                l=mid;            else                r=mid;        }        int q=0;        for(int i=0; i<m; i++)        {            if(q++)                printf(" ");            printf("%d",p[i].id);        }        printf("\n");    }    return 0;}