快来！Python详细解法

Python解法：

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.count = n  # 初始连通块数量为n（每个元素独立）
    
    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])  # 路径压缩
        return self.parent[x]
    
    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x != root_y:
            self.parent[root_x] = root_y  # 合并集合
            self.count -= 1  # 连通块数量减1

def count_beacon_networks(n, beacons):
    # 蜂巢网格的6个相邻方向
    directions = [(-1, -1), (-1, 0), (0, -1), (0, 1), (1, 0), (1, 1)]
    
    # 用字典存储激活单元，键为坐标元组，值为唯一ID（0到n-1）
    beacon_map = {tuple(beacon): i for i, beacon in enumerate(beacons)}
    uf = UnionFind(n)
    
    # 遍历每个激活单元，检查其6个相邻位置
    for (x, y), idx in beacon_map.items():
        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if (nx, ny) in beacon_map:
                uf.union(idx, beacon_map[(nx, ny)])
    
    return uf.count

# 读取输入
n = int(input())
beacons = []
for _ in range(n):
    x, y = map(int, input().split())
    beacons.append((x, y))

# 输出结果
print(count_beacon_networks(n, beacons))