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