poj K Best 最大化平均值 二分搜索

K Best

Time Limit: 8000MS Memory Limit: 65536KTotal Submissions: 6315 Accepted: 1689Case 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

分析：Σvi / Σwi>=x，要求得最大x，上述公式可以变形为 Σvi-x*Σwi>=0; 因此二分遍历x的取值即可得到符合的结果；

code：

<span style="font-size:18px;">#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<vector>using namespace std;const int Maxn=100001;const double eps=1e-8;struct node{int w,v,index;double r;};node sn[Maxn];int n,k;bool cmp(node x,node y){return x.r>y.r;}bool can(double mid){for(int i=0;i<n;i++){sn[i].r=sn[i].v-mid*sn[i].w;}//sort(sn,sn+n,cmp);partial_sort(sn,sn+k,sn+n,cmp);double sum=0;for(int i=0;i<k;i++){sum+=sn[i].r;}return sum>=0;}int main(){while(cin>>n>>k){int sum=0;for(int i=0;i<n;i++){scanf("%d %d",&sn[i].v,&sn[i].w);sn[i].index=i+1;sum+=sn[i].v;}double lb=0.0,ub=sum*1.0;while(fabs(ub-lb)>eps){double mid=(lb+ub)/2;if(can(mid)) lb=mid;else ub=mid;}for(int i=0;i<k-1;i++){cout<<sn[i].index<<" ";}cout<<sn[k-1].index<<endl;}return 0;} </span>