BZOJ3198:[SDOI2013]SPRING

Description

Input

Output

Sample Input

3 3
1 2 3 4 5 6
1 2 3 0 0 0
0 0 0 4 5 6

Sample Output

2

HINT

题解：

一脸容斥的样子。

枚举判断是否相同的泉水集合S，若|S|>=K，则inc(ANS,(-1)^(|S|-K)*C(|S|,K)*相同对数)。

用哈希表记录、判断（我之前竟然写了类似字符串哈希的的做法，哈希值相同直接判断相同，WA惨了）。

代码：

var
  i,j,k,l,n,m:longint;
  f:array[0..200001,1..6]of int64;
  b:array[1..6]of longint;
  c:array[0..6,0..6]of int64;
  cc:array[0..133330]of longint;
  d:array[0..200001,-1..6]of longint;
  a:array[0..200001]of int64;
  ans:int64;
procedure find(x,z:longint;y:int64);
var i,j,l:longint;
begin
  i:=cc[a[z]];
  while i>0 do
  begin
    l:=0;
    for j:=1 to x do if d[i,j]<>f[z,b[j]] then begin l:=1; break; end;
    if l=0 then begin ans:=ans+y*d[i,0]; inc(d[i,0]); exit; end;
    i:=d[i,-1];
  end;
  inc(m); for i:=1 to x do d[m,i]:=f[z,b[i]];
  d[m,0]:=1; d[m,-1]:=cc[a[z]]; cc[a[z]]:=m;
end;
function ss2(x:longint;y:int64):int64;
var i,j,k:longint;
begin
  ss2:=0; m:=0;
  for i:=0 to 133330 do cc[i]:=0;
  for i:=1 to n do
  begin
    a[i]:=1;
    for j:=1 to x do a[i]:=(a[i]*13131+f[i,b[j]])mod 133331;
    find(x,i,y);
  end;
end;
procedure ss(x,y:longint);
var i:longint;
begin
  if x>6 then
  begin
    if y>=k then
    begin
      if(y-k)mod 2=0 then ss2(y,c[y,k])
      else ss2(y,-c[y,k]);
    end;
    exit;
  end;
  ss(x+1,y);
  inc(y); b[y]:=x; ss(x+1,y);
end;
begin
  c[0,0]:=1;
  for i:=1 to 6 do
  begin
    c[i,0]:=1; c[i,i]:=1;
    for j:=1 to i-1 do c[i,j]:=c[i-1,j-1]+c[i-1,j];
  end;
  readln(n,k);
  for i:=1 to n do
  for j:=1 to 6 do read(f[i,j]);
  ss(1,0);
  writeln(ans);
end.