POJ3111 K Best 2017-05-11 18:12 31人阅读 评论(0) 收藏

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 10261		Accepted: 2644
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

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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 long

int 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;
}