poj3111 K Best 【0-1分数规划】

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 7154		Accepted: 1875
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

题意：给你N个珠宝已经每个珠宝的价值和重量，你只能取其中K个。问你取哪几个珠宝使sigma(v[i])/ sigma(w[i])最大。

方法一：构造函数 sigma(t[i]) = sigma(v[i]) - sigma(w[i]) * o。 然后 二分o值

这道题没有那么高精度，用整型就足够了。 我用的double跑了6s。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define MAXN 100000+10
#define eps 1e-8
using namespace std;
struct Node
{
    double v, w, t;
    int id;
};
Node num[MAXN];
int N, K;
bool cmp(Node a, Node b)
{
    return a.t > b.t;
}
bool judge(double o)
{
    for(int i = 0; i < N; i++)
        num[i].t = num[i].v - o * num[i].w;
    sort(num, num+N, cmp);
    double sum = 0;
    for(int i = 0; i < K; i++)//取前K个最大的
        sum += num[i].t;
    return sum >= 0;
}
int main()
{
    while(scanf("%d%d", &N, &K) != EOF)
    {
        double r = 0;
        for(int i = 0; i < N; i++)
        {
            scanf("%lf%lf", &num[i].v, &num[i].w);
            num[i].id = i + 1;
            r = max(r, num[i].v / num[i].w);
        }
        double l = 0;
        while(r - l >= eps)
        {
            double mid = (l + r) / 2;
            if(judge(mid))
                l = mid;
            else
                r = mid;
        }
        for(int i = 0; i < K; i++)
        {
            if(i > 0) printf(" ");
            printf("%d", num[i].id);
        }
        printf("
");
    }
    return 0;
}