poj 3111 二分求最大化平均值

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原创最新推荐文章于 2026-03-08 00:05:50 发布 712 阅读 AI 写同款· GEO 优化 ›

MIMO中K-Best仿真算法 MIMO中K-Best仿真算法，64QAM调制，发射天线数与接收天线数均为8（可调整），可绘出信噪比曲线立即下载

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 6169		Accepted: 1651
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

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

这道题就是用贪心+二分求最大化平均值。

选前k个的时候用nth_element比用sort快多了...

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

#define maxn 100005



int ans[maxn];
int n, k;
double x;

struct node
{
    int v, w;
    int id;

    bool operator < (const node &a) const
    {
        return v-x*w > a.v-x*a.w;
    }
};
node a[maxn];

int main()
{
    while(scanf("%d%d", &n, &k)!=EOF){
        for(int i = 0; i < n; i++){
            scanf("%d%d", &a[i].v, &a[i].w);
            a[i].id = i+1;
        }

        double l = 0,  r = 1e7+10;
        while(l+1e-6<r){
            double m = (r+l)/2;
            x = m;

            nth_element(a, a+k-1, a+n);
            double tmp= 0;
            for(int i = 0; i < k; i++)
                tmp += a[i].v-x*a[i].w;

            if(tmp+1e-5 >= 0){
                for(int i = 0; i < k; i++)
                    ans[i] = a[i].id;
                l = m;
            }
            else
                r = m;
        }

        for(int i = 0; i < k-1; i++)
            printf("%d ", ans[i]);
        printf("%d\n", ans[k-1]);

    }

    return 0;
}