POJ3111 K Best 2017-05-11 18:12 31人阅读 评论(0) 收藏

K Best Time Limit: 8000MS   Memory Limit: 65536K Total Submissions: 10261   Accepted: 2644 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 . 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 Source Northeastern Europe 2005, Northern Subregion

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题目的的意思是给出n个分数取其中m个，要求分子之和与分母之和之比最大

思路：二分+验证

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <vector>
using namespace std;
#define inf 0x3f3f3f3f
#define LL long long

int n,m;
struct node
{
    int a,b,id;
} p[100005];
double k;

bool cmp(node a,node b)
{
    double x=a.a-(a.b*k);
    double y=b.a-(b.b*k);
    return x>y;
}

bool ok(double mid)
{
    k=mid;
    sort(p,p+n,cmp);
    LL x=0,y=0;
    for(int i=0; i<m; i++)
    {
        x+=p[i].a;
        y+=p[i].b;
    }
    double ans=x*1.0/y;
    if(ans>=mid)
        return 1;
    return 0;

}

int main()
{
    while(~scanf("%d%d",&n,&m)&&(m||n))
    {
        for(int i=0; i<n; i++)
            scanf("%d%d",&p[i].a,&p[i].b),p[i].id=i+1;
        double l=0,r=1e6;
        while(r-l>1e-5)
        {
            double mid=(l+r)/2;
            if(ok(mid))
                l=mid;
            else
                r=mid;
        }
        int q=0;
        for(int i=0; i<m; i++)
        {
            if(q++)
                printf(" ");
            printf("%d",p[i].id);
        }
        printf("\n");
    }
    return 0;
}