POJ_3111_K_Best_(二分,最大化平均值)

描述

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http://poj.org/problem?id=3111

n个珠宝,第i个有价值v[i]和重量w[i],要从总选k个,使得这k个的(价值之和)/(重量之和)即平均价值最大,输出选中的珠宝编号.

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 8189		Accepted: 2087
Case 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 2
1 1
1 2
1 3

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

分析

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二分.

最大化平均值(同POJ 2976 Dropping Tests).

不等式变形做就可以了.

注意:

1.算c的值的时候不用担心算爆,int会先被提升为double再做运算,所以r取INF也没关系,不会爆(最多是INF*INF).

2.但是r取INF精度就不够了,所以还是乖乖取max(a[i]/b[i])把...

3.排序的时候直接排成从大到小的,方便.

4.这题二分次数少了进度不够,次数多了超时. 

#include<cstdio>
#include<algorithm>
using std :: sort;
using std :: max;

const int maxn=100005,INF=0x7fffffff;
int n,k;
int ans[maxn];
struct point
{
    int a,b,num;
    double c;
}j[maxn];
double x,y;

bool comp(point x,point y) { return x.c>y.c; }

bool C(double x)
{
    for(int i=1;i<=n;i++) j[i].c=j[i].a-j[i].b*x;
    sort(j+1,j+n+1,comp);
    double sum=0;
    for(int i=1;i<=k;i++) sum+=j[i].c;
    return sum>=0;
}

void solve()
{
    for(int i=0;i<25;i++)
    {
        double m=x+(y-x)/2;
        if(C(m))
        {
            x=m;
            for(int i=1;i<=k;i++) ans[i]=j[i].num;
        }
        else y=m;
    }
    for(int i=1;i<k;i++) printf("%d ",ans[i]);
    printf("%d",ans[k]);
}

void init()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++) 
    {
        scanf("%d%d",&j[i].a,&j[i].b);
        j[i].num=i;
        y=max(y,(double)j[i].a/(double)j[i].b);
    }
}

int main()
{
    freopen("k.in","r",stdin);
    freopen("k.out","w",stdout);
    init();
    solve();
    fclose(stdin);
    fclose(stdout);
    return 0;
}