最大化平均值---二分搜索 - R_xiaozhu_Q的专栏 - 博客频道 - CSDN.NET


有n个物品的重量和价值分别是w[i]和v[i],从中选出K个物品使得单位重量的价值最大。
1<=k<=n<=10^4
1<=w[i],v[i]<=10^6
一般想到的是按单位价值对物品排序,然后贪心选取,但是这个方法是错误的,对于有样例不满足。我们一般用二分搜索来做(其实这就是一个01分数规划)
我们定义:
条件 C(x) :=可以选k个物品使得单位重量的价值不小于x。
因此原问题转换成了求解满足条件C(x)的最大x。那么怎么判断C(x)是否满足?
变形:(sigma(v[i])/sigma(w[i]))>=x (i 属于我们选择的某个物品集合S)
进一步:sigma(v[i]-x*w[i])>=0
于是:条件满足等价于选最大的k个和不小于0.于是排序贪心选择可以判断,每次判断的复杂度是O(nlogn)。
  1. const double eps=1e-5;  
  2. int w[maxn],v[maxn],n,k;  
  3. double y[maxn];  
  4. bool check(double r)  
  5. {  
  6.     for(int i=0;i<n;i++){  
  7.         y[i]=v[i]-r*w[i];  
  8.     }  
  9.     sort(y,y+n);  
  10.     reverse(y,y+n);  
  11.     double sum=0;  
  12.     for(int i=0;i<k;i++)  
  13.     {  
  14.         sum+=y[i];  
  15.     }  
  16.     return sum>=0;  
  17. }   
  18. int main()  
  19. {  
  20.     while(cin>>n)  
  21.     {  
  22.         cin>>k;  
  23.         for(int i=0;i<n;i++)  
  24.             cin>>w[i]>>v[i];  
  25.         double lb=0,ub=1e6;  
  26.         while(ub-lb>eps)  
  27.         {  
  28.             double mid=(lb+ub)/2;  
  29.             if(check(mid)) lb=mid;  
  30.             else ub=mid;  
  31.         }  
  32.         printf("%.2f\n",ub);       
  33.     }  
  34.     return 0;  
  35. }
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