POJ-3111 K Best

题目链接

https://vjudge.net/problem/POJ-3111

题面

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 = \({i_1, i_2, …, i_k}\) as
\[ s(S)=\frac{\sum_\limits{j=1}^kv_{i_j}}{\sum_\limits{j=1}^kw_{i_j}} \]

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 ≤ kn ≤ 100 000).

The following n lines contain two integer numbers each — vi and wi (0 ≤ vi\(10^6\), 1 ≤ wi\(10^6\), both the sum of all vi and the sum of all wi do not exceed \(10^7\)).

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

给定n个数对\((v_i,w_i)\),必须选择k个数,问他们的s最小是多少,s的定义见题面

题解

这题不太容易想到是二分,但其实这是个二分答案的题,我们先二分一个s,那么我们可以化简一下下面这个式子
\[ s(S)=\frac{\sum_\limits{j=1}^kv_{i_j}}{\sum_\limits{j=1}^kw_{i_j}} \]
变成
\[ s(S) \times \sum_\limits{j=1}^kw_{i_j}=\sum_\limits{j=1}^kv_{i_j} \]
那么对于\(i_1...i_k\)都可以得到一个\(v_{i_j}-s(S)\times w_{i_J}\),显然s(S)越大这个值越小,那么我们就找出我们把这个值存进一个数组中从大到小排一下序,选取k个,如果能大于0的话说明s(S)可以更大,小于零则不行,实数二分一下即可,最后输出选取的k个数即可

AC代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#define N 100050
#define eps 1e-8
using namespace std;
typedef long long ll;
int n, k;
struct node {
    double v, w, tmp;
    int id;
} a[N];
bool cmp(node a, node b) {
    return a.tmp > b.tmp;
}
bool check(double x) {
    for (int i = 1; i <= n; i++) {
        a[i].tmp = a[i].v - a[i].w * x;
    }
    sort(a + 1, a + n + 1, cmp);
    double nowsum = 0;
    for (int i = 1; i <= k; i++) {
        nowsum += a[i].tmp;
    }
    return nowsum >= 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;
    }
    double l = 0, r = 1e6;
    double mid;
    while(r - l > eps) {
        mid = (l + r) / 2;
        if (check(mid)) {
            l = mid;
        }
        else r = mid;
    }
    for (int i = 1; i <= k; i++) {
        printf("%d ", a[i].id);
    }
    return 0;
}

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转载自www.cnblogs.com/artoriax/p/10375709.html
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