POJ

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 11571 Accepted: 2983Case 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

题意：有n个珠宝的价值和重量分别是vi和wi，从中选出k个珠宝，使得 ∑v / ∑w 最大，求使得其最大的所选的k个珠宝

假设∑v / ∑w ≥ x 则问题便转化成 找到最大的使得不等式 ∑v / ∑w ≥ x 成立的 x

∑v / ∑w ≥ x  = > ∑v ≥ x * ∑w = ∑ (x*w)  = > ∑ (v - x * w) ≥  0

利用二分查找求得最大x

#include<iostream>#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn = 100000+100;int n,k,v[maxn],w[maxn],s[maxn];struct node{    double s;int pos;}a[maxn];bool cmp(node a,node b) {return a.s>b.s;}int main(){    scanf("%d%d",&n,&k);    for(int i=0;i<n;i++) scanf("%d%d",&v[i],&w[i]);    double l = 0,r = 1e7,mid = (l+r)/2;    // > 1e-6 时会有精度误差    while(1e-6<r-l){        mid = (l+r)/2;        for(int i=0;i<n;i++){            a[i].s = v[i] - mid * w[i];            a[i].pos = i+1;        }        sort(a,a+n,cmp);        double sum = 0;        for(int i=0;i<k;i++) sum += a[i].s;        if(sum>=0) l = mid;        else r = mid;    }    for(int i=0;i<k;i++){        if(i) printf(" ");        printf("%d",a[i].pos);    }    printf("\n");    return 0;}