PKU 3111 K Best

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 = {i ₁ , i ₂ , …, 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

这道题的迭代方法我真是不会，希望能通过作越来越多的题突破瓶颈，sigh

 

假设已有的v/w，其中v和w是k个的总和。那么要得到更好的v/w，做法是，求出每一个对于的si，si＝ vi-wi*v/w

 

 

很容易就能证明si更大的k个，可以组成不比原来小的组合。 设前k个si之和为s;(s最大，如果s>0，迭代没有停止)，v/w = (v1-s)/w1<=v1/w1;

 

//#include "stdafx.h"
//============================================================================
// Name        : acmcpp.cpp
// Author      : Yubo Chow
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <algorithm>
#include <set>
#include <limits.h>
#include <float.h>
#include <bitset>
#include <cmath>
#include <vector>
#include <map>
#include <list>
using namespace std;

int n, k;

typedef struct
{
	int w, v;
} jewel;

jewel jewels[100001];

int num[100001];
double s[100001];

inline bool cmp(const int a, const int b)
{
	return s[a] > s[b];
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("case.txt", "r", stdin);
#endif

	while (scanf("%d%d", &n, &k) == 2)
	{

		for (int i = 0; i < n; ++i)
		{
			scanf("%d%d", &jewels[i].v, &jewels[i].w);
			num[i] = i;
		}

		int w = 0, v = 0;

		for (int i = 0; i < k; ++i)
		{
			v += jewels[i].v;
			w += jewels[i].w;
		}

		double ret;
		do
		{
			ret = double(v) / double(w);

			for (int i = 0; i < n; ++i)
			{
				s[i] = (double) jewels[i].v - ret * jewels[i].w;
			}
			nth_element(num, num + k, num + n, cmp);

			v = w = 0;
			for (int i = 0; i < k; ++i)
			{
				v += jewels[num[i]].v;
				w += jewels[num[i]].w;
			}

		} while (double(v) / double(w) - ret > 0.0000000001);

		//sort(num, num + k);
		printf("%d", num[0] + 1);
		for (int i = 1; i < k; ++i)
		{
			printf(" %d", num[i] + 1);
		}
		printf("/n");
	}

	return 0;
}