POJ 3111

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 11555 Accepted: 2978Case 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：

就是最大化平均值。

POINT：

二分答案。

二分搜索法模型：    条件C(x):可以挑选使得单位重量的物品价值不小于x->求满足条件的最大x->如何判断C(x)    价值和/重量和>=x    价值和-重量和*x>=0    和（价值-重量*x）>=0    可以对（价值-重量*x）的值进行贪心的选取，选取最大的k个 和>=0

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>#include <math.h>using namespace std;const int maxn = 100100;typedef pair<double, int> pr;int n,k;int v[maxn];int w[maxn];int check(double x){    pr y[maxn];    for(int i=1;i<=n;i++){        y[i].first=v[i]-x*w[i];        y[i].second=i;    }    sort(y+1,y+1+n);    double sum=0;    for(int i=n;i>=n-k+1;i--){        sum+=y[i].first;    }    if(sum>0) return 1;    return 0;}void haha(double x){    pr y[maxn];    for(int i=1;i<=n;i++){        y[i].first=v[i]-x*w[i];        y[i].second=i;    }    sort(y+1,y+1+n);    for(int i=n;i>=n-k+1;i--){        if(i!=n) printf(" ");        printf("%d",y[i].second);    }    printf("\n");}void solve(){    double l=0,r=100000;    while(fabs(l-r)>1e-6){        double mid=(l+r)/2;        if(check(mid)){            l=mid;        }else{            r=mid;        }    }    haha(l);}int main(){    while(~scanf("%d %d",&n,&k)){        for(int i=1;i<=n;i++){            scanf("%d %d",&v[i],&w[i]);        }        solve();    }}