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[0..101,0..32768] of double;
    v,a:array[0..20] 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:=1 shl k-1;
  for i:=1 to k do
  begin
    read(v[i]);
    read(x);
    while x<>0 do
    begin
      a[i]:=a[i]+1 shl (x-1);
      read(x);
    end;
  end;
  for i:=n downto 1 do
  begin
    for j:=0 to m do
    begin
      for p:=1 to k do
        if (a[p] and j)=a[p] then
          f[i,j]:=f[i,j]+max(f[i+1,j],f[i+1,j or (1 shl (p-1))]+v[p])
        else f[i,j]:=f[i,j]+f[i+1,j];
      f[i,j]:=f[i,j]/k;
    end;
  end;
  writeln(f[1,0]:0:6);
end.