bzoj1076

好像题目有点问题？且现在决定不吃的宝物以后也不能再吃
好像下次还是可以吃的，就是这样
然后明显是状压dp,注意这道题明显倒推更容易
因为顺推很可能从一个无效状态推到有效状态

 var f:array[..,..] of double;
     v,a:array[..] of longint;
     i,j,n,m,p,x,k:longint;

 function max(a,b:double):double;
   begin
     if a>b then exit(a) else exit(b);
   end;

 begin
   readln(n,k);
   m:= shl k-;
   for i:= to k do
   begin
     read(v[i]);
     read(x);
     while x<> do
     begin
       a[i]:=a[i]+ shl (x-);
       read(x);
     end;
   end;
   for i:=n downto  do
   begin
     for j:= to m do
     begin
       for p:= to k do
         if (a[p] and j)=a[p] then
           f[i,j]:=f[i,j]+max(f[i+,j],f[i+,j or ( shl (p-))]+v[p])
         else f[i,j]:=f[i,j]+f[i+,j];
       f[i,j]:=f[i,j]/k;
     end;
   end;
   writeln(f[,]::);
 end.