poj 3111 K Best 二分搜索 最大化平均值

K Best

Time Limit:		Memory Limit:
Total Submissions:		Accepted:
Case Time Limit:		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

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

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

Sample Input

3 21 1 1 2 1 3

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

错因分析：不能写成 sigma(v)/sigma(w)<=x

分析：挑战上143页，二分搜索，记sigma(v)/sigma(w)>=x,问题转换成了求x得最大值，二分搜索就行

• #include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
int num[105];
struct Node{
int w,v,id;
double temp;
}node[100005];
bool cmp(Node a,Node b)
{
return a.temp>b.temp;
}
int n,k;
int ok(double x)
{
double sum=0;
for(int i=1;i<=n;i++)
        node[i].temp=node[i].v-node[i].w*x;
sort(node+1,node+1+n,cmp);
for(int i=1;i<=k;i++)
        sum+=node[i].temp;
return sum>=0;
}
int main()
{
while(~scanf("%d %d",&n,&k))
{
for(int i=1;i<=n;i++)
{
scanf("%d %d",&node[i].v,&node[i].w);
            node[i].id=i;
}
double l=0,mid,r=1e6;
for(int i=1;i<=40;i++)  //开成100次会超时，20次会wa，可能有poj服务器的问题
{
         mid=l+(r-l)/2;
if(ok(mid))
             l=mid;
else
             r=mid;
}
for(int i=1;i<=k;i++)
printf("%d ",node[i].id);
printf("\n");
}
return 0;
}