codeforces 739E

官方题解是一个n2logn的dp做法

不过有一个简单易想的费用流做法

对每个小精灵，连边(A,i,1,pi) (B,i,1,ui) (i,t,1,0) (i,t,1,-pi*ui)

最后连边(s,A,a,0) (s,B,b,0)

跑最大费用最大流即可，注意精度误差

 #include<bits/stdc++.h>

 using namespace std;

 struct way{int po,next,flow;double cost;} e[];
 const double eps=1e-;
 int pre[],p[],cur[],q[];
 double d[],c[];
 bool v[];
 int len,n,m,a,b,t;

 void add(int x,int y,int f,double c)
 {
      e[++len].po=y;
      e[len].flow=f;
      e[len].cost=c;
      e[len].next=p[x];
      p[x]=len;
 }

 void build(int x,int y,int f,double c)
 {
      add(x,y,f,c);
      add(y,x,,-c);
 }

 bool spfa()
 {
      for (int i=; i<=t; i++) d[i]=-1e20;
      memset(v,,sizeof(v));
      d[]=;
      int f=,r=;q[]=;
      while (f<=r)
      {
            int x=q[f++];
            v[x]=;
            for (int i=p[x]; i!=-; i=e[i].next)
            {
                int y=e[i].po;
                if (e[i].flow&&d[x]+e[i].cost-eps>d[y])
                {
                   d[y]=d[x]+e[i].cost;
                   pre[y]=x; cur[y]=i;
                   if (!v[y])
                   {
                      q[++r]=y;
                      v[y]=;
                   }
                }
            }
      }
      return d[t]>-1e20;
 }

 double cost()
 {
     double s=;
     while (spfa())
     {
           s+=d[t];
           for (int i=t; i; i=pre[i])
           {
               int j=cur[i];
               e[j].flow--;
               e[j^].flow++;
           }
     }
     return s;
 }

 int main()
 {
     len=-;
     memset(p,,sizeof(p));
     scanf("%d%d%d",&n,&a,&b);
     t=n+;
     for (int i=; i<=n; i++)
     {
         scanf("%lf",&c[i]);
         build(n+,i,,c[i]);
         build(i,t,,);
     }
     for (int i=; i<=n; i++)
     {
         double x;
         scanf("%lf",&x);
         build(n+,i,,x);
         build(i,t,,-c[i]*x);
     }
     build(,n+,a,);
     build(,n+,b,);
     printf("%.5lf\n",cost());
 }