POJ3111-K Best

最新推荐文章于 2019-08-10 16:34:00 发布

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解决Demy面临的难题，从n个具有不同价值和重量的珠宝中选出k个，使这k个珠宝的平均价值最大化。使用二分查找算法确定最佳组合。

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K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 15077		Accepted: 3859
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

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

代码：

/*
    题意：求n个东西里取k个，使它们平均值最大，输出这个k个东西序号。 
*/
#include "iostream"
#include "cstdio"
#include "cstring"
#include "cmath"
#include "algorithm"
int n,k;
const int maxn = 1000000;
using namespace std;
struct jew
{
    int id;
    double value;
    bool operator<(const jew&a) const{
        if(value>a.value) return true;
        return false;
    }
} jews[maxn+10];
double v[maxn+10],w[maxn+10];
bool check(double mid)
{
    double sum=0;
    for(int i=1; i<=n; i++)
    {
        jews[i].id=i;
        jews[i].value=v[i]-w[i]*mid;
    }
    sort(jews+1,jews+n+1);
    for(int i=1; i<=k; i++)
    {
        sum+=jews[i].value;
    }
    return sum>=0;
}
void solve()
{
    double lb= 0, lr = maxn;
    //cout<<lb<<" "<<lr<<endl;
    while(fabs(lr-lb)>=1e-6)
    {
        double mid = (lb + lr) / 2;
        if (check(mid)) lb = mid;
        else lr = mid;
    }
    for(int i=1; i<=k; i++)
    {
        if(i==k)
            printf("%d",jews[i].id);
        else
            printf("%d ",jews[i].id);
    }

}
int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%lf %lf",&v[i],&w[i]);
            //jews[i].value=v[i]*1.0/w[i];
        }
    solve();printf("\n");
    }
    return 0;
}

思考：这道题作为上一道题目Dropping Test的延续，上一道是删去贡献小的成绩，提高平均成绩而这道题是选择k个，使得平均值最大。

题目WA了好几发，后来发现用POJ的C++编译器可以过！！！

还是一样的二分算法，不再赘述