AtCoder Regular Contest 083 - Collecting Balls

There are 2N balls in the xy-plane. The coordinates of the i-th of them is (x_i, y_i).
Here, x_i and y_i are integers between 1 and N (inclusive) for all i, and no two balls occupy the same coordinates.
In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B.
Then, he placed the type-A robots at coordinates (1, 0), (2, 0), ..., (N, 0), and the type-B robots at coordinates (0, 1), (0, 2), ..., (0, N), one at each position.
When activated, each type of robot will operate as follows.
 - When a type-A robot is activated at coordinates (a, 0), it will move to the position of the ball with the lowest y-coordinate among the balls on the line x = a, collect the ball and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.
 - When a type-B robot is activated at coordinates (0, b), it will move to the position of the ball with the lowest x-coordinate among the balls on the line y = b, collect the ball and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything.
Once deactivated, a robot cannot be activated again. Also, while a robot is operating, no new robot can be activated until the operating robot is deactivated.
When Snuke was about to activate a robot, he noticed that he may fail to collect all the balls, depending on the order of activating the robots.
Among the (2N)! possible orders of activating the robots, find the number of the ones such that all the balls can be collected, modulo 1 000 000 007.

-----Constraints-----
 - 2 \leq N \leq 10^5
 - 1 \leq x_i \leq N
 - 1 \leq y_i \leq N
 - If i ≠ j, either x_i ≠ x_j or y_i ≠ y_j.

-----Inputs-----
Input is given from Standard Input in the following format:
N
x_1 y_1
...
x_{2N} y_{2N}

-----Outputs-----
Print the number of the orders of activating the robots such that all the balls can be collected, modulo 1 000 000 007.

-----Sample Input-----
2
1 1
1 2
2 1
2 2

-----Sample Output-----
8

We will refer to the robots placed at (1, 0) and (2, 0) as A1 and A2, respectively, and the robots placed at (0, 1) and (0, 2) as B1 and B2, respectively.
There are eight orders of activation that satisfy the condition, as follows:
 - A1, B1, A2, B2
 - A1, B1, B2, A2
 - A1, B2, B1, A2
 - A2, B1, A1, B2
 - B1, A1, B2, A2
 - B1, A1, A2, B2
 - B1, A2, A1, B2
 - B2, A1, B1, A2
Thus, the output should be 8.

Difficulty: VERY_HARD

Starter code:

Solutions:
import sys
input = sys.stdin.readline
MOD = 10 ** 9 + 7
N = int(input())
ball = (tuple((int(x) for x in row.split())) for row in sys.stdin.readlines())
graph = [set() for _ in range(N + N + 1)]
for (x, y) in ball:
	graph[x].add(y + N)
	graph[y + N].add(x)
visited = [False] * (N + N + 1)
components = []
for x in range(1, N + N + 1):
	if visited[x]:
		continue
	V = set([x])
	E = []
	q = [x]
	visited[x] = True
	while q:
		y = q.pop()
		for z in graph[y]:
			if y < z:
				E.append((y, z))
			if visited[z]:
				continue
			V.add(z)
			visited[z] = True
			q.append(z)
	components.append((V, E))

def make_get_pattern(V):
	deg1 = [x for x in V if len(graph[x]) == 1]
	get = {}
	while deg1:
		x = deg1.pop()
		if not graph[x]:
			continue
		y = graph[x].pop()
		se = graph[y]
		se.remove(x)
		if len(se) == 1:
			deg1.append(y)
		if x < y:
			get[x, y] = 0
		else:
			get[y, x] = 1
	for x in V:
		if graph[x]:
			y = graph[x].pop()
			break
	graph[y].remove(x)
	if x > y:
		(x, y) = (y, x)
	get[x, y] = 2
	while graph[x]:
		y = graph[x].pop()
		graph[y].remove(x)
		if x < y:
			get[x, y] = 3
		else:
			get[y, x] = 2
		x = y
	return get

def F(V, E):
	if len(E) != len(V):
		return 0
	ret = 0
	E.sort()
	get = make_get_pattern(V)
	(den1, den2) = (1, 1)
	dp1 = {x: 0 for x in V}
	dp2 = {x: 0 for x in V}
	for (x, y) in E:
		if get[x, y] == 0:
			k1 = dp1[x] + 1
			k2 = dp2[x] + 1
		elif get[x, y] == 1:
			k1 = dp1[y] + 1
			k2 = dp2[y] + 1
		elif get[x, y] == 2:
			k1 = dp1[x] + 1
			k2 = dp2[y] + 1
		else:
			k1 = dp1[y] + 1
			k2 = dp2[x] + 1
		dp1[x] += k1
		dp1[y] += k1
		dp2[x] += k2
		dp2[y] += k2
		den1 *= k1
		den2 *= k2
		den1 %= MOD
		den2 %= MOD
	return sum((pow(x, MOD - 2, MOD) for x in (den1, den2)))
prob = 1
for c in components:
	prob *= F(*c)
	prob %= MOD
answer = prob
for n in range(1, N + N + 1):
	answer *= n
	answer %= MOD
print(answer)


import sys
input = sys.stdin.readline
MOD = 10 ** 9 + 7
N = int(input())
ball = [tuple((int(x) for x in input().split())) for _ in range(N + N)]
graph = [set() for _ in range(N + N + 1)]
for (x, y) in ball:
	graph[x].add(y + N)
	graph[y + N].add(x)
visited = set()
components = []
for x in range(1, N + N + 1):
	if x in visited:
		continue
	V = set([x])
	q = [x]
	while q:
		y = q.pop()
		for z in graph[y]:
			if z in V:
				continue
			V.add(z)
			q.append(z)
	visited |= V
	components.append(V)

def make_get_patterns(V):
	deg1 = [x for x in V if len(graph[x]) == 1]
	get = ({}, {})
	while deg1:
		x = deg1.pop()
		if not graph[x]:
			continue
		y = graph[x].pop()
		se = graph[y]
		se.remove(x)
		if len(se) == 1:
			deg1.append(y)
		if x < y:
			get[0][x, y] = 0
			get[1][x, y] = 0
		else:
			pass
			get[0][y, x] = 1
			get[1][y, x] = 1
	for x in V:
		if graph[x]:
			y = graph[x].pop()
			break
	graph[y].remove(x)
	if x > y:
		(x, y) = (y, x)
	get[0][x, y] = 0
	get[1][x, y] = 1
	while True:
		if not graph[x]:
			break
		y = graph[x].pop()
		graph[y].remove(x)
		if x < y:
			get[0][x, y] = 1
			get[1][x, y] = 0
		else:
			get[0][y, x] = 0
			get[1][y, x] = 1
		x = y
	return get

def F(V):
	E = sorted(((x, y) for x in V if x <= N for y in graph[x]))
	if len(E) != len(V):
		return 0
	ret = 0
	for get in make_get_patterns(V):
		den = 1
		dp = {x: 0 for x in V}
		for (x, y) in E:
			if get[x, y] == 0:
				k = dp[x] + 1
			else:
				k = dp[y] + 1
			dp[x] += k
			dp[y] += k
			den *= k
			den %= MOD
		ret += pow(den, MOD - 2, MOD)
	return ret % MOD
prob = 1
for c in components:
	prob *= F(c)
	prob %= MOD
answer = prob
for n in range(1, N + N + 1):
	answer *= n
	answer %= MOD
print(answer)