POJ 3111 K Best (01分数规划+二分)

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 6658		Accepted: 1756
Case 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 2
1 1
1 2
1 3

Sample Output

1 2

题目大意不再说了，很简单。

01分数规划，和poj2976一样，具体的不多说。求sum(vi)/sum(wi)=x，转化得：sum(vi)-sum(wi)*x=0，所以二分x即可。

注意代码中的v和w不要单独定义 double v[maxn],w[maxn]

然后：

struct pp
{
	double c;
	int id;
}s[maxn];

这样去写会WA。

然后注意精度，50就可以，100会TLE。

/*
01分数规划+记录选择
*/
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define eps 1e-10

const int maxn=100000+10;
int k,n;

struct pp
{
	double c;
	int v,w;	//v，w一定要和id同在一个结构体里，不然会WA死
	int id;
}s[maxn];


int cmp(const pp &x,const pp &y){
	return x.c>y.c;
}

int cmp1(const pp &x,const pp &y){
	return x.id<y.id;
}

bool C(double x){
	int i;
	double sum=0;
	for(i=0;i<n;i++){
		s[i].c=s[i].v-s[i].w*x;
	}
	sort(s,s+n,cmp);
	for(i=0;i<k;i++)
		sum+=s[i].c;
	if(sum<0) return false;
	return true;
}

int main()
{
	int i;
	while(scanf("%d%d",&n,&k)!=EOF){
		for(i=0;i<n;i++){
			scanf("%d%d",&s[i].v,&s[i].w);
			s[i].id=i+1;
		}
		double l=0,r=10000000;
		for(i=0;i<50;i++){
			double m=(l+r)/2;
			if(C(m))
				l=m;
			else
				r=m;
		}
		sort(s,s+k,cmp1);
		for(i=0;i<k;i++)
			printf("%d%c",s[i].id,i==k-1?'\n':' ');
	}
	return 0;
}