poj3111 二分最大化平均值

如题：http://poj.org/problem?id=3111

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 7073		Accepted: 1848
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 2
1 1
1 2
1 3

Sample Output

1 2

Source
Northeastern Europe 2005, Northern Subregion






思路：选取k个最大化平均值的思路在我上一篇，这里不再赘述。在不断缩小区间的过程中，满足最后一个精度的前k个元素也就是结果。循环100次超时，循环50次6秒。



#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

#define MAXN 100005

struct node

{

int v,w,id;

};

node a[MAXN];

struct node1

{

double value;

int id;

}b[MAXN];

bool operator<(node1 x,node1 y)

{

return x.value>y.value;

}

int C(double x,int k,int n)

{

int i;

for(i=0;i<n;i++)

{

b[i].value=a[i].v-x*a[i].w;

b[i].id=a[i].id;

}

sort(b,b+n);

double sum=0;

for(i=0;i<k;i++)

{

sum+=b[i].value;

}

return sum>=0;

}

int cmp(node1 x,node1 y)

{

return x.id<y.id;

}

int main()

{

// freopen("C:\\1.txt","r",stdin);

int n,k;

cin>>n>>k;

int i;

double l=0,r=0;

for(i=0;i<n;i++)

{

cin>>a[i].v>>a[i].w;

a[i].id=i+1;

r+=a[i].v;

}

for(i=0;i<50;i++)

{

double mid=(l+r)/2;

if(C(mid,k,n))

l=mid;

else

r=mid;

}

sort(b,b+k,cmp);

for(i=0;i<k;i++)

{

printf("%d%c",b[i].id,i==k-1?'\n':' ');

}

return 0;

}