POJ:3111-K Best

KBest问题求解

最新推荐文章于 2021-08-31 22:32:30 发布

原创最新推荐文章于 2021-08-31 22:32:30 发布 · 144 阅读

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#poj #二分平均值

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本文介绍了一个经典的KBest问题，并提供了一种通过二分搜索来寻找最大特定价值的珠宝组合的解决方案。面对财务危机，主人公Demy需要选择价值最大的k个珠宝保留。文章通过具体的输入输出示例解释了如何实现这一目标。

K Best

Time Limit: 8000MS Memory Limit: 65536K
Total Submissions: 12985 Accepted: 3324
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 sum(vi)/sum(wi)

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

解题心得：

还是一个二分平均值的问题，不懂得可以去看看挑战程序设计上面关于二分的推导。

#include <algorithm>
#include <stdio.h>
using namespace std;
const int maxn = 1e6+100;
struct Jewels {
    int va,w,num;
    double c;
}jew[maxn];

int n,k;

void init() {
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++) {
        scanf("%d%d",&jew[i].va,&jew[i].w);
        jew[i].num = i;
    }
}

bool cmp(Jewels a, Jewels b) {
    return a.c > b.c;
}

bool checke(double ave) {
    double sum = 0;
    for(int i=1;i<=n;i++) {
        jew[i].c = (double)jew[i].va - (double)jew[i].w*ave;
    }
    sort(jew+1,jew+n+1,cmp);
    for(int i=1;i<=k;i++) {
        sum += jew[i].c;
    }
    return sum >= 0;
}

void binary_search() {
    double l,r;
    r = 1000000000.0,l = 0;
    for(int i=0;i<50;i++) {
        double mid = (l + r) / 2;
        if(checke(mid))
            l = mid;
        else
            r = mid;
    }
}

int main() {
    init();
    binary_search();
    for(int i=1;i<=k;i++)
        printf("%d ",jew[i].num);
    return 0;
}