poj 3111 K Best 二分搜索 最大化平均值

K Best

Time Limit: 8000MSMemory Limit: 65536KTotal Submissions: 7623Accepted: 1970Case Time Limit: 2000MSSpecial 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

Source

Northeastern Europe 2005, Northern Subregion

错因分析：不能写成sigma(v)/sigma(w)<=x

分析：挑战上143页，二分搜索，记sigma(v)/sigma(w)>=x,问题转换成了求x得最大值，二分搜索就行

#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int num[105];
struct Node{
   int w,v,id;
   double temp;
}node[100005];
bool cmp(Node a,Node b)
{
    return a.temp>b.temp;
}
int n,k;
int ok(double x)
{
    double sum=0;
    for(int i=1;i<=n;i++)
        node[i].temp=node[i].v-node[i].w*x;
    sort(node+1,node+1+n,cmp);
    for(int i=1;i<=k;i++)
        sum+=node[i].temp;
    return sum>=0;
}
int main()
{
   while(~scanf("%d %d",&n,&k))
   {
      for(int i=1;i<=n;i++)
        {
            scanf("%d %d",&node[i].v,&node[i].w);
            node[i].id=i;
        }
      double l=0,mid,r=1e6;
      for(int i=1;i<=40;i++)  //开成100次会超时，20次会wa，可能有poj服务器的问题
      {
         mid=l+(r-l)/2;
         if(ok(mid))
             l=mid;
         else
             r=mid;
      }
      for(int i=1;i<=k;i++)
        printf("%d ",node[i].id);
     printf("\n");
   }
   return 0;
}