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POJ 3111 二分
| Time Limit: 8000MS | Memory Limit: 65536K | |
| Total Submissions: 10507 | Accepted: 2709 | |
| Case Time Limit: 2000MS | Special Judge |
Description
Demy has n jewels. Each of her jewels has some value vi and weight wi.
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, …, ik} 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 — vi and wi (0 ≤ vi ≤ 106, 1 ≤ wi ≤ 106, both the sum of all vi and the sum of all wi 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
//二分题，设最终结果是s,sum(分子)/sum(分母)=s,组成s的必然有a/b>=s和a/b<=s =>//a-s*b>=0和a-s*b<=0，因此二分s值然后按照a-s*b从大到小排序依次取前k个看结果是否//大于等于s#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100009;const double esp=0.000001;int n,k;double x;struct Lu{ int id; double v,w; bool operator < (const Lu &p)const{ return v-x*w>p.v-x*p.w; }}L[maxn];bool solve(double m){ x=m; sort(L,L+n); double tmp1=0,tmp2=0; for(int i=0;i<k;i++){ tmp1+=L[i].v; tmp2+=L[i].w; } return tmp1-m*tmp2>=0.0;}int main(){ while(scanf("%d%d",&n,&k)==2){ for(int i=0;i<n;i++){ scanf("%lf%lf",&L[i].v,&L[i].w); L[i].id=i; } double l=0.0,r=10000000.0; while(r-l>esp){ double m=(l+r)/2.0; if(solve(m)){ l=m; }else r=m; } for(int i=0;i<k;i++) printf("%d%c",L[i].id+1,i==k-1?‘\n‘:‘ ‘); } return 0;}
POJ 3111 二分