poj3111 选取物品（二分+贪心）

题目传送门

题意就是，n个物品，从中选取k个，要按照那个公式所取的值最大。

思路：最大化平均值的时候首先想到的就是二分， 我们设G(x) 为单位的重量不小于X， 我们的目标就是要找到满足条件的最大的X， 也就是说我们的OK函数就要判断是否能够找到一个大小为K的集合S， 使得单位重量的价值大于等于X， 我们将当时变形转化为vi - x*wi >= 0  对于 所有的i 属于集合S， 这时我们对于一个整体值vi-x*wi的值的 排序， 贪心的选取前K个， 这样OK函数就写好了。

但是这道题最恶心的地方是，卡常数，如果要用c++交，你的排序函数就必须 引用，g++交好像就没关系。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<math.h>
#include<cmath>
#include<time.h>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<algorithm>
#include<numeric>
using namespace std;
typedef long long ll;
const int maxn=2000010;
const int INF=0x3f3f3f3f;
struct dian{
	double v,w;
	double y;
	int id;
}a[maxn];
double mps=1e10,eps=1e-9;
double l,r,mid;
int n,k;
bool cmp(const dian& a,const dian& b){
	return a.y>b.y;
}
bool ok(double d){
		double e=0;
		for(int i=1;i<=n;i++){
			a[i].y=a[i].v-a[i].w*d;
		}
		sort(a+1,a+1+n,cmp);
		for(int i=1;i<=k;i++){
			e+=a[i].y;
		}
		return e>=0;
}
int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%lf%lf",&a[i].v,&a[i].w);
		a[i].id=i;
	}
	l=0,r=1e13;
	while(r-l>eps){
		mid=(l+r)/2;
		if(ok(mid)){
			l=mid;
		}else{
			r=mid;
		}
	}
	for(int i=1;i<=k;i++){
		printf("%d ",a[i].id);
	}
	printf("
");
}

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 13263		Accepted: 3388
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