"""
有 n 个城市，其中一些彼此相连，另一些没有相连。如果城市 a 与城市 b 直接相连，且城市 b 与城市 c 直接相连，那么城市 a 与城市 c 间接相连。
省份 是一组直接或间接相连的城市，组内不含其他没有相连的城市。
给你一个 n x n 的矩阵 isConnected ，其中 isConnected[i][j] = 1 表示第 i 个城市和第 j 个城市直接相连，而 isConnected[i][j] = 0 表示二者不直接相连。
返回矩阵中 省份 的数量。
示例 1：
输入：isConnected = [[1,1,0],[1,1,0],[0,0,1]]
输出：2
示例 2：
输入：isConnected = [[1,0,0],[0,1,0],[0,0,1]]
输出：3
"""
import collections
def findCircleNum(isConnected: list) -> int:
'''
深度优先
:param isConnected:
:return:
'''
def dfs(i: int):
for j in range(cities):
if isConnected[i][j] == 1 and j not in visited:
visited.add(j)
dfs(j)
cities = len(isConnected)
visited = set()
provinces = 0
for i in range(cities):
if i not in visited:
dfs(i)
provinces += 1
return provinces
def findCircleNum1(isConnected: list) -> int:
'''
广度优先
:param isConnected:
:return:
'''
cities = len(isConnected)
visited = set()
provinces = 0
for i in range(cities):
if i not in visited:
Q = collections.deque([i])
while Q:
j = Q.popleft()
visited.add(j)
for k in range(cities):
if isConnected[j][k] == 1 and k not in visited:
Q.append(k)
provinces += 1
return provinces
def findCircleNum2(isConnected: list) -> int:
'''
并查集
:param isConnected:
:return:
'''
def find(index: int) -> int:
if parent[index] != index:
parent[index] = find(parent[index])
return parent[index]
def union(index1: int, index2: int):
parent[find(index1)] = find(index2)
cities = len(isConnected)
parent = list(range(cities))
for i in range(cities):
for j in range(i + 1, cities):
if isConnected[i][j] == 1:
union(i, j)
provinces = sum(parent[i] == i for i in range(cities))
return provinces
print(findCircleNum1([[1,1,0],[1,1,0],[0,0,1]]))
