POJ

 

一、题目概述

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 12167		Accepted: 3126
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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二、题目释义

有N颗珠宝，每颗珠宝的价值为vi，重量为wi。 女主不得已要卖掉部分珠宝，她想留下k颗珠宝，并要求(v1+v2+....vk) / (w1+w2+...wk)的值最大，输出女主留下的珠宝的编号。（可不按输入的顺序输出）。

三、思路分析

四、AC代码

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 1e5+5;
const double esp = 1e-6;

struct jewel
{
    double v,w,t;
    int id;
}S[N];
int n,k;
bool cmp(jewel a, jewel b)
{
    return a.t > b.t;
}

bool check(double x)
{
    double sum = 0;
    for(int i=0; i<n; i++)
    {
        S[i].t = S[i].v - x*S[i].w;
    }
    sort(S,S+n,cmp);
    for(int i=0; i<k; i++)
    {
        sum += S[i].t;
    }
    return sum >= 0;
}

int main()
{
    double l,r;
    double mid;

    while(scanf("%d%d",&n,&k) != EOF)
    {
        r = 0;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf",&S[i].v,&S[i].w);
            S[i].id = i+1;
            r = max(r,S[i].v / S[i].w);
        }
        while(r-l>esp)
        {
            mid = (l+r) / 2;
            if(check(mid)) l = mid;
            else r = mid;
        }
        for(int i=0; i<k-1; i++)
        {
            cout << S[i].id << " ";
        }
        cout << S[k-1].id << endl;
    }
    return 0;
}