1688: [Usaco2005 Open]Disease Manangement 疾病管理

1688: [Usaco2005 Open]Disease Manangement 疾病管理

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 413  Solved: 275
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Description

Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) different diseases among them, then the milk will be too contaminated and will have to be discarded in its entirety. Please help determine the largest number of cows FJ can milk without having to discard the milk.

Input

* Line 1: Three space-separated integers: N, D, and K * Lines 2..N+1: Line i+1 describes the diseases of cow i with a list of 1 or more space-separated integers. The first integer, d_i, is the count of cow i's diseases; the next d_i integers enumerate the actual diseases. Of course, the list is empty if d_i is 0. 有N头牛,它们可能患有D种病,现在从这些牛中选出若干头来,但选出来的牛患病的集合中不过超过K种病.

Output

* Line 1: M, the maximum number of cows which can be milked.

Sample Input

6 3 2
0---------第一头牛患0种病
1 1------第二头牛患一种病,为第一种病.
1 2
1 3
2 2 1
2 2 1

Sample Output

5

OUTPUT DETAILS:

If FJ milks cows 1, 2, 3, 5, and 6, then the milk will have only two
diseases (#1 and #2), which is no greater than K (2).

HINT

 

Source

Silver

题解：其实一开始想的很复杂，连树状数组都想过用上，但是后来发现貌似也就15种病，这样子就容易了——直接进行01状态压缩，然后枚举各种状态（HansBug：实际上，对于疾病的状态只需要保留刚刚好K种病即可，就算是更少的病种也可以剪掉，想想为什么^_^），然后就是各种位运算，然后A掉么么哒

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 var
    i,j,k,l,m,n,x,ans:longint;
    a:array[..] of longint;
 function min(x,y:longint):longint;
          begin
               if x<y then min:=x else min:=y;
          end;
 function max(x,y:longint):longint;
          begin
               if x>y then max:=x else max:=y;
          end;
 begin
      readln(n,m,l);l:=min(l,m);ans:=;
      for i:= to n do
          begin
               a[i]:=;
               read(k);
               for j:= to k do
                   begin
                        read(x);
                        a[i]:=a[i] or ( shl (x-));
                   end;
               readln;
          end;
      for i:= to trunc(exp(ln()*m)-) do
          begin
               j:=i;k:=;
               while j> do
                     begin
                          inc(k,j mod );
                          j:=j div ;
                     end;
               if k<>l then continue;k:=;
               for j:= to n do if (i or a[j])=i then inc(k);
               ans:=max(ans,k);
          end;
      writeln(ans);
 end.