Problem:
There are N cities. There are also K roads and L railways, extending between the cities. The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities. No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.  We will say city A and B are _connected by roads_ if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads. We will also define _connectivity by railways_ similarly.  For each city, find the number of the cities connected to that city by both roads and railways.

Incorrect submission (verdict: Wrong Answer):
from collections import deque

N, K, L = [int(s) for s in input().split()]
edge_road = [[int(s) - 1 for s in input().split()] for _ in range(K)]
edge_train = [[int(s) - 1 for s in input().split()] for _ in range(L)]
graph_road = [[] for _ in range(N)]
graph_train = [[] for _ in range(N)]

for i, j in edge_road:
    graph_road[i].append(j)
    graph_road[j].append(i)

for i, j in edge_train:
    graph_train[i].append(j)
    graph_train[j].append(i)

completed = [False] * N

road = {}
train = {}

group_road = [0] * N
group_num = 0

visited = [False]*N

for i in range(N):
    if not visited[i]:
        continue
    dq = deque([i])
    while(dq):
        search_obj = dq.pop()
        for candidate_obj in graph_road[search_obj]:
            if not visited[candidate_obj]:
                group_road[candidate_obj] = group_num
                dq.appendleft(search_obj)
                visited[candidate_obj] = True
    group_num += 1


group_train = [0] * N
group_num = 0

visited = [False]*N

for i in range(N):
    if not visited[i]:
        continue
    dq = deque([i])
    while(dq):
        search_obj = dq.pop()
        for candidate_obj in graph_train[search_obj]:
            if not visited[candidate_obj]:
                group_train[candidate_obj] = group_num
                dq.appendleft(search_obj)
                visited[candidate_obj] = True
    group_num += 1

group_count = {}

for gr, gt in zip(group_road, group_train):
    if (gr, gt) in group_count.keys():
        group_count[(gr, gt)] += 1
    else:
        group_count[(gr, gt)] = 1


for gr, gt in zip(group_road, group_train):
    print(group_count[(gr, gt)] - 1, end=' ')


Fixed, accepted submission:
N, M, L = [int(s) for s in input().split()]
parent_road = list(range(N))
parent_train = list(range(N))
rank_road = [0] * N
rank_train = [0] * N


class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * N

    def find(self, x):
        if x == self.parent[x]:
            return x
        else:
            self.parent[x] = self.find(self.parent[x])
            return self.parent[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.parent[x] = y
        else:
            self.parent[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1


edge_road = [[int(s) - 1 for s in input().split()] for _ in range(M)]
edge_train = [[int(s) - 1 for s in input().split()] for _ in range(L)]

group_road = UnionFind(N)
group_train = UnionFind(N)

for x, y in edge_road:
    group_road.union(x, y)

for x, y in edge_train:
    group_train.union(x, y)

group_count = {}
group_list = []
for i in range(N):
    gr = group_road.find(i)
    gt = group_train.find(i)
    group_list.append((gr, gt))
    if (gr, gt) in group_count.keys():
        group_count[(gr, gt)] += 1
    else:
        group_count[(gr, gt)] = 1

print(' '.join([str(group_count[g]) for g in group_list]))