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count-valid-paths-in-a-tree.py
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count-valid-paths-in-a-tree.py
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# Time: O(n)
# Space: O(n)
# number theory, tree dp, iterative dfs
class Solution(object):
def countPaths(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: int
"""
def linear_sieve_of_eratosthenes(n): # Time: O(n), Space: O(n)
primes = []
spf = [-1]*(n+1) # the smallest prime factor
for i in xrange(2, n+1):
if spf[i] == -1:
spf[i] = i
primes.append(i)
for p in primes:
if i*p > n or p > spf[i]:
break
spf[i*p] = p
return spf
def is_prime(u):
return spf[u] == u
def iter_dfs():
result = 0
stk = [(1, (0, -1, [0]*2))]
while stk:
step, args = stk.pop()
if step == 1:
u, p, ret = args
ret[:] = [1-is_prime(u+1), is_prime(u+1)]
stk.append((2, (u, p, ret, 0)))
elif step == 2:
u, p, ret, i = args
if i == len(adj[u]):
continue
v = adj[u][i]
stk.append((2, (u, p, ret, i+1)))
if v == p:
continue
new_ret = [0]*2
stk.append((3, (u, p, new_ret, ret, i)))
stk.append((1, (v, u, new_ret)))
elif step == 3:
u, p, new_ret, ret, i = args
result += ret[0]*new_ret[1]+ret[1]*new_ret[0]
if is_prime(u+1):
ret[1] += new_ret[0]
else:
ret[0] += new_ret[0]
ret[1] += new_ret[1]
return result
spf = linear_sieve_of_eratosthenes(n)
adj = [[] for _ in xrange(n)]
for u, v in edges:
u, v = u-1, v-1
adj[u].append(v)
adj[v].append(u)
return iter_dfs()
# Time: O(n)
# Space: O(n)
# number theory, tree dp, dfs
class Solution2(object):
def countPaths(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: int
"""
def linear_sieve_of_eratosthenes(n): # Time: O(n), Space: O(n)
primes = []
spf = [-1]*(n+1) # the smallest prime factor
for i in xrange(2, n+1):
if spf[i] == -1:
spf[i] = i
primes.append(i)
for p in primes:
if i*p > n or p > spf[i]:
break
spf[i*p] = p
return spf
def is_prime(u):
return spf[u] == u
def dfs(u, p):
cnt = [1-is_prime(u+1), is_prime(u+1)]
for v in adj[u]:
if v == p:
continue
new_cnt = dfs(v, u)
result[0] += cnt[0]*new_cnt[1]+cnt[1]*new_cnt[0]
if is_prime(u+1):
cnt[1] += new_cnt[0]
else:
cnt[0] += new_cnt[0]
cnt[1] += new_cnt[1]
return cnt
spf = linear_sieve_of_eratosthenes(n)
adj = [[] for _ in xrange(n)]
for u, v in edges:
u, v = u-1, v-1
adj[u].append(v)
adj[v].append(u)
result = [0]
dfs(0, -1)
return result[0]
# Time: O(n)
# Space: O(n)
# number theory, union find
class UnionFind(object): # Time: O(n * alpha(n)), Space: O(n)
def __init__(self, n):
self.set = range(n)
self.rank = [0]*n
self.size = [1]*n
def find_set(self, x):
stk = []
while self.set[x] != x: # path compression
stk.append(x)
x = self.set[x]
while stk:
self.set[stk.pop()] = x
return x
def union_set(self, x, y):
x, y = self.find_set(x), self.find_set(y)
if x == y:
return False
if self.rank[x] > self.rank[y]: # union by rank
x, y = y, x
self.set[x] = self.set[y]
if self.rank[x] == self.rank[y]:
self.rank[y] += 1
self.size[y] += self.size[x]
return True
def total(self, x):
return self.size[self.find_set(x)]
class Solution3(object):
def countPaths(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: int
"""
def linear_sieve_of_eratosthenes(n): # Time: O(n), Space: O(n)
primes = []
spf = [-1]*(n+1) # the smallest prime factor
for i in xrange(2, n+1):
if spf[i] == -1:
spf[i] = i
primes.append(i)
for p in primes:
if i*p > n or p > spf[i]:
break
spf[i*p] = p
return spf
def is_prime(u):
return spf[u] == u
spf = linear_sieve_of_eratosthenes(n)
uf = UnionFind(n)
for u, v in edges:
u, v = u-1, v-1
if is_prime(u+1) == is_prime(v+1) == False:
uf.union_set(u, v)
result = 0
cnt = [1]*n
for u, v in edges:
u, v = u-1, v-1
if is_prime(u+1) == is_prime(v+1):
continue
if not is_prime(u+1):
u, v = v, u
result += cnt[u]*uf.total(v)
cnt[u] += uf.total(v)
return result