poj 3111 K Best（二分）

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 9341 Accepted: 2423Case 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 21 11 21 3

Sample Output

1 2

平均值最大化问题

题意：有n个物品的重量和价值分别是wi和vi。从中选出k个物品使得单位重量的价值最大
也就是  k个物体的w相加除以k个物体的v相加最大

一般想到的是按单位价值对物品排序，然后贪心选取，但是这个方法是错误的，比如对nyoj的例题来说，从大到小地进行选取，输入的结果是5/7=0.714对于有样例不满足。
我们一般用二分搜索来做（其实这就是一个01分数规划）


我们定义：


条件 C(x) :=可以选k个物品使得单位重量的价值不小于x。


因此原问题转换成了求解满足条件C(x)的最大x。那么怎么判断C(x)是否满足？


变形：（sigma(v[i])/sigma(w[i])）>=x (i 属于我们选择的某个物品集合S)


进一步：sigma(v[i]-x*w[i])>=0


于是：条件满足等价于选最大的k个和不小于0.于是排序贪心选择可以判断，每次判断的复杂度是O(nlogn)。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 100005;const double E = 1e-6;int n ,k,v[maxn],w[maxn];struct node{    int id;    double x;}a[maxn];int cmp(node a ,node b){    return a.x > b.x;}void bs(double l , double r){    while(r-l > E)              //以为除以肯定x（m）不一定是整数，所以用double    {        double m = (l+r)/2;        double ans = 0,ansl = 0;        for(int i = 1 ; i <= n;i++)        {            a[i].x = v[i] - m*w[i];            a[i].id = i;            //一定要有这个，因为原来那个顺序一直没变，变得只是一个临时数组a[i]；所以用id记录他的位置，然后sort排序，让最大的前n个放前面        }        sort(a+1,a+1+n,cmp);        for(int i = 1; i <= k;i++)        {            ans += a[i].x;        }        if(ans >= 0)  {ansl = m; l = m;}        else  r = m;    }    return ;}int main(){    while(cin >> n >> k)    {        for(int i = 1; i <= n;i++)        {            cin >> v[i] >> w[i] ;        }        bs(0,maxn);        int flag = 0;        for(int i = 1; i <= k ; i++)         {            if(flag++)  cout << ' ';            cout << a[i].id;         }         cout << endl;    }    return 0;}