POJ3111

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

题解：

AC代码为：

 个题很明显就是要最大化平均值，然而采用贪心的方法每次取单位价值最大的钻石，是显然不行的。所以应该采用二分搜索的方法，那么二分什么值最后才能得到答案呢？不妨这样想，S（x）表示最终的平均价值，那么就相当于找到一组（v,w）组合使得Σv/Σw≥S(x)，移项得到不等式Σ（v-S(x)*w）≥0，这样一来就很容易发现直接二分S(x)即可，每次得到一个S(x)计算v-S(x)*w，之后排序，取前k个计算是否大于等于0，知道二分到一个S(x)使得不等式的值为0就是答案。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <cstring>
#include <string>
#include <map>
#include <queue>
#include <set>


using namespace std;


const int maxn = 100000 + 5;
const double inf = 1000000 + 5;
int n,k;
struct jew 
{
    int id;
    int v,w;
    double key;
    void cal(double x) 
{ 
key = v - x * w; 
}

    bool operator < (const jew& a) const 
{
        return key > a.key;
    }
}j[maxn];


bool c(double x) 
{
    for (int i = 1; i <= n; i++) 
{
        j[i].cal(x);
    }
    
    sort(j+1,j+n+1);
    
    double tmp = 0;
    for (int i = 1; i <= k; i++) 
tmp += j[i].key;
    return tmp >= 0;
}


int main() 
{
    while (cin>>n>>k) 
{
        for (int i = 1; i <= n; i++) 
{
            scanf("%d%d",&j[i].v,&j[i].w);
            j[i].id = i;
        }
        double l = 0, r = inf;
        for (int i = 0; i < 100; i++) 
{
            double mid = (l + r) / 2;
            if (c(mid)) 
l = mid;
            else 
r = mid;
        }
        for (int i = 1; i <= k; i++) 
{
            if (i - 1) 
printf(" %d",j[i].id);
            else 
printf("%d",j[i].id);
        }
        
        cout<<endl;
    }
}