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

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 = ; i <= n; ++i) {

                 f[i] = (v[i] - x * w[i]);

         }

         sort(f + ,f + n + );

         double sum = ;

         for(int i = n; i >= n - k + ; --i) {

                 sum += f[i];

         }

         return sum >= ;

 }

 void output(double x) {

         for(int i = ; i <= n; ++i) {

                 ans[i] = make_pair(v[i] - x * w[i],i);

         }

         sort(ans + ,ans + n + );

         printf("%d",ans[n].second);

         for(int i = n - ; i >= n - k + ; --i) {

                 printf(" %d",ans[i].second);

         }

         printf("\n");

 }

 void solve() {

         double l = ,r = ;

         for(int i = ; i <= n; ++i) {

                 r += v[i];

         }

         int num = ;

         while(num++ <  ) {

                 //printf(" l = %f r = %f\n",l,r);

                 //num++;

                 double mid = (l + r) / ;

                 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 = ; i <= n; ++i) {

             scanf("%d%d",&v[i],&w[i]);

         }

         solve();

     }

     return ;

 }

巴特西

POJ 3111

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