POJ-3111 K Best---二分求最大化平均值

题目链接：

https://cn.vjudge.net/problem/POJ-3111

题目大意：

卖宝救夫：Demy要卖珠宝，n件分别价值v_i 重 w_i，她希望保留k件使得

最大。

解题思路：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;
 const int INF = 1e7;
 const int maxn = ;
 int n, m;
 double w[maxn], v[maxn], y[maxn];
 bool judge(double x)
 {
     for(int i = ; i <= n; i++)
         y[i] = v[i] - x * w[i];
     sort(y + , y + n + );

     double sum = ;
     for(int i = , j = n; i <= m; i++, j--)
         sum += y[j];
     return sum >= ;
 }
 struct node
 {
     double x;
     int id;
     bool operator <(const node& a)const
     {
         return x > a.x;
     }
 }ans[maxn];
 int main()
 {
     cin >> n >> m;
     for(int i = ; i <= n; i++)
         cin >> v[i] >> w[i];
     double l = , r = INF;
     for(int i = ; i < ; i++)
     {
         double mid = (l + r) / ;
         if(judge(mid))l = mid;
         else r = mid;
     }
     for(int i = ; i <= n; i++)
     {
         ans[i].id = i;
         ans[i].x = v[i] - l * w[i];
     }
     sort(ans + , ans + n + );
     printf("%d", ans[].id);
     for(int i = ; i <= m; i++)
     {
         printf(" %d", ans[i].id);
     }
     puts("");
     return ;
 }