POJ 3111 K best 二分

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

Sample Output

1 2

正常的01分数规划

跟上一篇文章一样https://mp.csdn.net/postedit/80602317

乱搞一下

输出即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
const int MAX_N = 1024;
int n,m;
double x;
struct node {
   int w;
   int s;
   bool operator < (const node& other) const{
      return w-x*s>other.w-x*other.s;
   }
}arr[MAX_N];
int check(double p){
    double total_a = 0, total_b = 0;
    x = p;
    sort(arr+1,arr+1+n);
    for(int i = 1;i<=(n-m);++i){
        total_a+=arr[i].w;
        total_b+=arr[i].s;
    }
    return total_a/total_b > p;
}
int main(){
 while(~scanf("%d%d",&n,&m)&&(n||m)){
    for(int i=1;i<=n;++i)
        scanf("%d",&arr[i].w);
    for(int i=1;i<=n;++i)
        scanf("%d",&arr[i].s);
    double l = 0,r=1;
    while(abs(r-l)>1e-4){
        double mid =(l+r)/2;
        if(check(mid)) l=mid;
        else r=mid;
    }
    printf("%.0f\n",l*100);
 }
 return 0;
}

耗时较久可以用牛顿迭代

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