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Tenzing and Random Real Numbers

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Description:

There are $$$n$$$ uniform random real variables between 0 and 1, inclusive, which are denoted as $$$x_1, x_2, \ldots, x_n$$$.

Tenzing has $$$m$$$ conditions. Each condition has the form of $$$x_i+x_j\le 1$$$ or $$$x_i+x_j\ge 1$$$.

Tenzing wants to know the probability that all the conditions are satisfied, modulo $$$998~244~353$$$.

Formally, let $$$M = 998~244~353$$$. It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{p}{q}$$$, where $$$p$$$ and $$$q$$$ are integers and $$$q \not \equiv 0 \pmod{M}$$$. Output the integer equal to $$$p \cdot q^{-1} \bmod M$$$. In other words, output the integer $$$x$$$ that $$$0 \le x < M$$$ and $$$x \cdot q \equiv p \pmod{M}$$$.

Input:

The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n\le 20$$$, $$$0\le m\le n^2+n$$$).

Then following $$$m$$$ lines of input contains three integers $$$t$$$, $$$i$$$ and $$$j$$$ ($$$t \in \{0,1\}$$$, $$$1\le i\le j\le n$$$).

  • If $$$t=0$$$, the condition is $$$x_i+x_j\le 1$$$.
  • If $$$t=1$$$, the condition is $$$x_i+x_j\ge 1$$$.

It is guaranteed that all conditions are pairwise distinct.

Output:

Output the probability that all the conditions are satisfied, modulo $$$M = 998~244~353$$$.

Sample Input:

3 2
0 1 2
1 3 3

Sample Output:

748683265

Sample Input:

3 3
0 1 2
0 1 3
0 2 3

Sample Output:

748683265

Sample Input:

3 4
0 1 2
0 1 3
1 2 3
1 2 2

Sample Output:

935854081

Sample Input:

4 4
0 1 2
0 3 4
1 1 3
1 2 4

Sample Output:

0

Sample Input:

8 12
0 1 2
0 2 3
1 3 4
0 1 4
0 5 6
0 6 7
1 7 8
0 5 8
1 3 7
1 3 8
1 4 7
1 4 8

Sample Output:

997687297

Note:

In the first test case, the conditions are $$$x_1+x_2 \le 1$$$ and $$$x_3+x_3\ge 1$$$, and the probability that each condition is satisfied is $$$\frac 12$$$, so the probability that they are both satisfied is $$$\frac 12\cdot \frac 12=\frac 14$$$, modulo $$$998~244~353$$$ is equal to $$$748683265$$$.

In the second test case, the answer is $$$\frac 14$$$.

In the third test case, the answer is $$$\frac 1{16}$$$.

In the fourth test case, the sum of all variables must equal $$$2$$$, so the probability is $$$0$$$.

Informação

Codeforces

Provedor Codeforces

Código CF1842H

Tags

bitmasksdpgraphsmathprobabilities

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Datas 09/05/2023 10:40:20

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