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):
#UNIONFIND
class UnionFind:
    def __init__(self, n):
        
        
        self.table = [-1] * n
 
    def _root(self, x):
        if self.table[x] < 0:
            return x
        else:
            
            self.table[x] = self._root(self.table[x])
            return self.table[x]
 
    def find(self, x, y):
        return self._root(x) == self._root(y)
 
    def union(self, x, y):
        r1 = self._root(x)
        r2 = self._root(y)
        if r1 == r2:
            return
        
        d1 = self.table[r1]
        d2 = self.table[r2]
        if d1 <= d2:
            self.table[r2] = r1
            if d1 == d2:
                self.table[r1] -= 1
        else:
            self.table[r1] = r2
 
 
if __name__ == '__main__':
    n,k,l=map(int,input().split())
    pq=UnionFind(n+1)
    rs=UnionFind(n+1)
    for i in range(k):
        p,q=map(int,input().split())
        pq.union(p,q)
    for i in range(l):
        r,s=map(int,input().split())
        rs.union(r,s)
    sameroot=[]
    for i in range(1,n+1):
        sameroot.append([pq._root(i),rs._root(i),i])
    sameroot.sort()
    sameroot.append([0,0,n+1])
    cnt=[0]*(n+1)
    bx,by=0,0
    samelist=[]
    for x,y,i in sameroot:
        if bx==x and by==y:
            samelist.append(i)
        else:
            k=len(samelist)
            if k>0:
                for j in samelist:
                    cnt[j]=k
            samelist=[i]
            bx,by=x,y
    print(cnt[1:])

Fixed, accepted submission:
#UNIONFIND
class UnionFind:
    def __init__(self, n):
        
        
        self.table = [-1] * n
 
    def _root(self, x):
        if self.table[x] < 0:
            return x
        else:
            
            self.table[x] = self._root(self.table[x])
            return self.table[x]
 
    def find(self, x, y):
        return self._root(x) == self._root(y)
 
    def union(self, x, y):
        r1 = self._root(x)
        r2 = self._root(y)
        if r1 == r2:
            return
        
        d1 = self.table[r1]
        d2 = self.table[r2]
        if d1 <= d2:
            self.table[r2] = r1
            if d1 == d2:
                self.table[r1] -= 1
        else:
            self.table[r1] = r2
 
 
if __name__ == '__main__':
    n,k,l=map(int,input().split())
    pq=UnionFind(n+1)
    rs=UnionFind(n+1)
    for i in range(k):
        p,q=map(int,input().split())
        pq.union(p,q)
    for i in range(l):
        r,s=map(int,input().split())
        rs.union(r,s)
    sameroot=[]
    for i in range(1,n+1):
        sameroot.append([pq._root(i),rs._root(i),i])
    sameroot.sort()
    sameroot.append([0,0,n+1])
    cnt=[0]*(n+1)
    bx,by=0,0
    samelist=[]
    for x,y,i in sameroot:
        if bx==x and by==y:
            samelist.append(i)
        else:
            k=len(samelist)
            if k>0:
                for j in samelist:
                    cnt[j]=k
            samelist=[i]
            bx,by=x,y
    print(' '.join(map(str,cnt[1:])))