poj3111 K Best【最大化平均值】

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 10485 Accepted: 2698Case 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

思路：二分解决最大化平均值，玄学的是C++超时，G++过了；

二分的是最后的平均值，将题目中式子变化一下就是 s[i].d = s[i].a - mid * s[i].b ;用out数组记录 i 就可以了；

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define max_n 100010using namespace std;typedef long long LL;double in[max_n];int out[max_n];int n,k;struct node{double a;double b;int c;double d;}s[max_n];bool cmp(node x,node y){return x.d>y.d;}bool judge(double mid){for(int i=0;i<n;i++){s[i].d=s[i].a-mid*s[i].b;}sort(s,s+n,cmp);double sum=0.0;for(int i=0;i<k;i++){sum+=s[i].d;out[i]=s[i].c;}return sum>=0.0;}int main(){while(~scanf("%d %d",&n,&k)){memset(s,0,sizeof(s));memset(out,0,sizeof(out));for(int i=0;i<n;i++){scanf("%lf %lf",&s[i].a,&s[i].b);s[i].c=i+1;}double l=0.0,r=10000000.0,mid;while(r-l>1e-6){mid=(l+r)/2.0;if(judge(mid)) l=mid;else r=mid;}//printf("%lf\n",l);for(int i=0;i<k;i++){if(i>0) printf(" ");printf("%d",out[i]); }  printf("\n");}return 0;}