POJ 3111 K Best(二分——最大化平均值)

K Best
Time Limit: 8000MS Memory Limit: 65536K
Total Submissions: 10457 Accepted: 2695
Case Time Limit: 2000MS Special Judge
Description

Demy has n jewels. Each of her jewels has some value vi and weight wi.

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, …, ik} 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 — vi and wi (0 ≤ vi ≤ 106, 1 ≤ wi ≤ 106, both the sum of all vi and the sum of all wi 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
Sample Output

1 2
Source

Northeastern Europe 2005, Northern Subregion

题意：有N个珠宝、选K个使这K个的单位质量的价值X最高

做法：对X二分，对于每个X我们选出vi-wi*x最大的K个，看是不是能>=0，如果能满足条件，说明x小了
为什么能这样做呢 x=∑v/∑ w 移项得
∑v-x*∑w=0=∑（vi-x*wi）这个值随着x的增大而减小，满足二分的单调性条件，所以可以二分做。

#include <iostream>#include <cmath>#include <cstdio>#include <cstring>#include <algorithm>#define maxn 100010using namespace std;int n,k;struct Node{    int id,v,w;    double a;}node[maxn];void setup(double x){    for(int i=0;i<n;i++)    {        node[i].a=node[i].v-node[i].w*x;    }}bool cmp(const Node &x,const Node & y){    return x.a>y.a;}bool judge(double x){    setup(x);    double ans=0;    sort(node,node+n,cmp);    for(int i=0;i<k;i++)    {        ans+=node[i].a;    }    return ans>=0;}int main(){    int a,b;    int minw=0;    int sumv=0;    while(scanf("%d%d",&n,&k)==2)    {        for(int i=0;i<n;i++)        {            scanf("%d%d",&a,&b);            node[i].id=i+1;            node[i].v=a;            node[i].w=b;        }        double st=0,ed=1e7,mid;        for(int i=0;i<100;i++)        {            mid=(st+ed)/2;            if(judge(mid))            {                st=mid;            }            else ed=mid;        }        for(int i=0;i<k;i++)        {            printf("%d ",node[i].id);        }        printf("\n");    }    return 0;}