POJ3111

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

题解：

AC代码为：

个题很明显就是要最大化平均值，然而采用贪心的方法每次取单位价值最大的钻石，是显然不行的。所以应该采用二分搜索的方法，那么二分什么值最后才能得到答案呢？不妨这样想，S（x）表示最终的平均价值，那么就相当于找到一组（v,w）组合使得Σv/Σw≥S(x)，移项得到不等式Σ（v-S(x)*w）≥0，这样一来就很容易发现直接二分S(x)即可，每次得到一个S(x)计算v-S(x)*w，之后排序，取前k个计算是否大于等于0，知道二分到一个S(x)使得不等式的值为0就是答案。

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <vector>

 #include <cstring>

 #include <string>

 #include <map>

 #include <queue>

 #include <set>

 using namespace std;

 const int maxn =  + ;

 const double inf =  + ;

 int n,k;

 struct jew

 {

     int id;

     int v,w;

     double key;

     void cal(double x)

 {

 key = v - x * w;

 }

     bool operator < (const jew& a) const

 {

         return key > a.key;

     }

 }j[maxn];

 bool c(double x)

 {

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

 {

         j[i].cal(x);

     }

     sort(j+,j+n+);

     double tmp = ;

     for (int i = ; i <= k; i++)

 tmp += j[i].key;

     return tmp >= ;

 }

 int main()

 {

     while (cin>>n>>k)

 {

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

 {

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

             j[i].id = i;

         }

         double l = , r = inf;

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

 {

             double mid = (l + r) / ;

             if (c(mid))

 l = mid;

             else

 r = mid;

         }

         for (int i = ; i <= k; i++)

 {

             if (i - )

 printf(" %d",j[i].id);

             else

 printf("%d",j[i].id);

         }

         cout<<endl;

     }

 }

巴特西

POJ3111

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