Cojs Strong connected Graph Count 1-2 report

OwO topic meaning is the same, but the data range expanded

For the n<=7 problem, we can do a direct brute force search.

For the n<=1000 problem, we can not be difficult to associate with < theme > This topic

Yes, you just have to change the equation a little.

First, we consider an illegal scheme. Strongly connected components must be a DAG after they are shrunk.

Consider child issues: Dag Count

Procedure can refer to the <cojs Dag Count 1-4 Solution Report >

This gives the transfer equation.

F (N) =sigma (( -1) ^ (k-1) *c (n,k) *2^ (k* (n-k)) *f (n-k))

What if we consider the case of a strong connected component's shrinking point?

We can define a case where g (n) indicates that n points can form several unconnected strong connected components

Consider calculating the tolerance coefficient, so if you make an odd number of strong connected components, the G (n) contribution is 1, otherwise 1

Defines the number of strongly connected graphs with n points for F (n)

Not hard to get recursive g (n) =f (n)-sigma (C (n-1,k-1) *f (k) *g (n-k))

The meaning is to consider the size of the strongly connected component of a point, the enumeration size K, then the selection scheme is C (n-1,k-1)

If the size is n, then the contribution is f (n)

Otherwise the contribution part is-g (n-k) *f (k), notice here because of a strong connected component

So the parity changes, G's positive and negative also change

Then we define h (n) to represent the number of graphs of n points, not difficult to get H (n) =2^ (n-1)

Considering the solution of F (n) by using the tolerance method, the analogy above solves the DAG and can get the following recursive

F (n) =h (n)-sigma (C (n,k) *2^ (k* (n-k)) *h (N-k) *g (k)) +f (n)

The meaning of this recursion is that a strongly connected component that enumerates some points is 0 after the indent.

Then the scheme in these points is the G (n) after the tolerance coefficient is calculated.

The selected point scheme is C (n,k), the remaining (n-k) points to the K-point arbitrary edge, the scheme is 2^ (k* (n-k))

The remaining (N-K) points are arbitrarily connected, and the scheme is H (n-k)

Notice here that the +f (n) is because when n=k, this lump is obtained by G (n), and we can only tolerate the solution of the strong connected component >1

So we get an O (n^2) recursive, can do n<=1000

For the n<=10w problem, we consider using FFT to optimize the above equation

In the above equation, F has g,g F, which is very disgusting.

Consider this equation f (n) =h (n)-sigma (C (n,k) *2^ (k* (n-k)) *h (N-k) *g (k)) +f (n)

Both sides f can be offset, we present a G (n) out and can get only about g of the recursive

g (n) =h (n)-sigma (C (n,k) *2^ (k* (n-k)) *h (N-k) *g (k)) (K-> (n-1))

about how 2^ (k* (n-k)) translates to the standard FFT form has been said in the DAG count.

namely nk=n^2/2+k^2/2-(N-K) ^2/2

And then do a cdq+fft, and we can find the G function.

then consider g (n) =f (n)-sigma (C (n-1,k-1) *f (k) *g (n-k))

Since the G function is known, we move the item to get

g (n) +sigma (C (n-1,k-1) *f (k) *g (n-k)) =f (n)

When we do a cdq+fft, we can find the F function.

Total time complexity O (nlog^2n)

Cojs Strong connected Graph Count 1-2 report