K Best
Time Limit: 8000MS   Memory Limit: 65536K
Total Submissions: 12812   Accepted: 3290
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
题目链接 点击打开链接

题目大意：给你N个珠宝已经每个珠宝的价值和重量，取其中K个。问取哪几个珠宝使得sigma(v[i])/ sigma(w[i])最大。
思路： 构造函数 sigma(t[i]) = sigma(v[i]) - sigma(w[i]) * x。 然后二分x值。   方法为0-1分数规划。 

#include<stdio.h>
#include<algorithm>
using namespace std;

#define exp 1e-8
struct jew{
	int id;
	double y;
}num[1000005];
double v[1000005], w[1000005];
bool cmp(jew a, jew b)
{
	return a.y > b.y;
}

bool dis(double x, int n, int k)
{
	int i;
	double sum = 0;
	for (i = 0; i < n; i++)
	{
		num[i].y = v[i] - x*w[i];
		num[i].id = i + 1;
	}
	sort(num, num + n, cmp);
	for (i = 0; i < k; i++)
	{
		sum += num[i].y;
	}
	return sum >= 0;
}
int main()
{
	int n, k;
	while (~scanf("%d %d", &n, &k))
	{
		for (int i = 0; i < n; i++)
		{
			scanf("%lf %lf", &v[i], &w[i]);
		}
		double left = 0, right = 1e6;
		double mid;
		while (right - left>=exp)
		{
			mid = (left + right) / 2;
			if (dis(mid, n, k))
			{
				left = mid;
			}
			else
			{
				right = mid;
			}
		}
		for (int i = 0; i < k - 1; i++)
		{
			printf("%d ", num[i].id);
		}
		printf("%d\n", num[k - 1].id);
	}



	return 0;
}