POJ 3111

K Best

Time Limit: 8000MS		Memory Limit: 65536K
Total Submissions: 5177		Accepted: 1411
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

3 2
1 1
1 2
1 3

Sample Output

1 2

Source

Northeastern Europe 2005, Northern Subregion

 

二分平均值

x为平均值，判断的时候 计算每个数的 (v[i] - w[i] * x) ，排序，找出最大的k个数，加起来，如果大于等于0 则往上找，否则往下

 

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

using namespace std;

#define maxn 100005
#define eps 1e-8
typedef pair<double,int> pii;

int w[maxn],v[maxn];
double f[maxn];
pii ans[maxn];
int n,k;


bool judge(double x) {
        for(int i = 1; i <= n; ++i) {
                f[i] = (v[i] - x * w[i]);
        }
        sort(f + 1,f + n + 1);

        double sum = 0;
        for(int i = n; i >= n - k + 1; --i) {
                sum += f[i];
        }

        return sum >= 0;

}

void output(double x) {

        for(int i = 1; i <= n; ++i) {
                ans[i] = make_pair(v[i] - x * w[i],i);
        }
        sort(ans + 1,ans + n + 1);

        printf("%d",ans[n].second);
        for(int i = n - 1; i >= n - k + 1; --i) {
                printf(" %d",ans[i].second);
        }

        printf("\n");


}

void solve() {
        double l = 0,r = 0;
        for(int i = 1; i <= n; ++i) {
                r += v[i];
        }

        int num = 0;
        while(num++ < 50 ) {
                //printf(" l = %f r = %f\n",l,r);
                //num++;
                double mid = (l + r) / 2;
                if(judge(mid)) {
                        l = mid;
                } else {
                        r = mid;
                }

        }

        //printf("%f\n",l);
        output(l);
}
int main()
{

    //freopen("sw.in","r",stdin);

    while(~scanf("%d%d",&n,&k))
    {
        for(int i = 1; i <= n; ++i) {
            scanf("%d%d",&v[i],&w[i]);
        }

        solve();
    }

    return 0;
}