Graph Subset Problem (Codeforces 1440D)
Tags: Others, hashing

You are given an undirected graph with $$$n$$$ vertices and $$$m$$$ edges. Also, you are given an integer $$$k$$$.Find either a clique of size $$$k$$$ or a non-empty subset of vertices such that each vertex of this subset has at least $$$k$$$ neighbors in the subset. If there are no such cliques and subsets report about it.A subset of vertices is called a clique of size $$$k$$$ if its size is $$$k$$$ and there exists an edge between every two vertices from the subset. A vertex is called a neighbor of the other vertex if there exists an edge between them.

Input: The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The next lines contain descriptions of test cases.The first line of the description of each test case contains three integers $$$n$$$, $$$m$$$, $$$k$$$ ($$$1 \leq n, m, k \leq 10^5$$$, $$$k \leq n$$$).Each of the next $$$m$$$ lines contains two integers $$$u, v$$$ $$$(1 \leq u, v \leq n, u \neq v)$$$, denoting an edge between vertices $$$u$$$ and $$$v$$$.It is guaranteed that there are no self-loops or multiple edges. It is guaranteed that the sum of $$$n$$$ for all test cases and the sum of $$$m$$$ for all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: For each test case: If you found a subset of vertices such that each vertex of this subset has at least $$$k$$$ neighbors in the subset in the first line output $$$1$$$ and the size of the subset. On the second line output the vertices of the subset in any order.If you found a clique of size $$$k$$$ then in the first line output $$$2$$$ and in the second line output the vertices of the clique in any order.If there are no required subsets and cliques print $$$-1$$$.If there exists multiple possible answers you can print any of them.

Note: In the first test case: the subset $$$\{1, 2, 3, 4\}$$$ is a clique of size $$$4$$$.In the second test case: degree of each vertex in the original graph is at least $$$3$$$. So the set of all vertices is a correct answer.In the third test case: there are no cliques of size $$$4$$$ or required subsets, so the answer is $$$-1$$$.

Sample Inputs:
3
5 9 4
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
10 15 3
1 2
2 3
3 4
4 5
5 1
1 7
2 8
3 9
4 10
5 6
7 10
10 8
8 6
6 9
9 7
4 5 4
1 2
2 3
3 4
4 1
1 3

Sample Outputs:
2
4 1 2 3 
1 10
1 2 3 4 5 6 7 8 9 10 
-1

Solution (Python):
from collections import deque
import sys
input = sys.stdin.readline
for _ in range(int(input())):
    (n, m, k) = map(int, input().split())
    G = [set() for _ in range(n + 1)]
    deg = [0] * (n + 1)
    for _ in range(m):
        (i, j) = map(int, input().split())
        G[i].add(j)
        G[j].add(i)
        deg[i] += 1
        deg[j] += 1
    if k * (k - 1) > 2 * m:
        print(-1)
        continue
    q = deque()
    seen = [0] * (n + 1)
    for i in range(1, n + 1):
        if deg[i] < k:
            seen[i] = 1
            q.append(i)
    clique = []
    end = 0
    while q and (not end):
        i = q.popleft()
        seen[i] = 2
        if deg[i] == k - 1 and (not clique) and (k * (k - 1) // 2 <= m):
            clique = [i]
            for j in G[i]:
                if seen[j] <= 1:
                    clique.append(j)
            ok = 1
            for x in clique:
                for y in clique:
                    if x == y:
                        break
                    if y not in G[x]:
                        ok = 0
                        break
                if not ok:
                    break
            if not ok:
                clique = []
            else:
                print(2)
                print(*clique)
                end = 1
                break
        for j in G[i]:
            deg[j] -= 1
            if deg[j] < k and (not seen[j]):
                seen[j] = 1
                q.append(j)
    if not end:
        subset = [i for i in range(1, n + 1) if not seen[i]]
        if subset:
            print('1 {}'.format(len(subset)))
            print(*subset)
        else:
            print(-1)

Solution (Python):
from __future__ import division, print_function
from heapq import *
import sys
le = sys.__stdin__.read().split('\n')[::-1]
af = []
for ezo in range(int(le.pop())):
    (n, m, k) = list(map(int, le.pop().split()))
    ar = [set() for i in range(n)]
    for i in range(m):
        (a, b) = map(int, le.pop().split())
        ar[a - 1].add(b - 1)
        ar[b - 1].add(a - 1)
    if k * (k - 1) > 2 * m:
        af.append(-1)
    else:
        pi = [i for i in range(n) if len(ar[i]) < k]
        re = set((i for i in range(n) if len(ar[i]) >= k))
        while pi:
            a = pi.pop()
            if len(ar[a]) == k - 1:
                clique = True
                for j in ar[a]:
                    if clique:
                        for h in ar[a]:
                            if not h in ar[j] and h != j:
                                clique = False
                                break
                if clique:
                    af.append(2)
                    af.append(' '.join(map(str, [i + 1 for i in ar[a]] + [a + 1])))
                    pi = []
                    re = {-1}
            for j in ar[a]:
                ar[j].remove(a)
                if len(ar[j]) < k:
                    if j in re:
                        pi.append(j)
                        re.discard(j)
        if re and re != {-1}:
            af.append('1' + ' ' + str(len(re)))
            af.append(' '.join(map(str, [i + 1 for i in re])))
        elif re != {-1}:
            af.append(-1)
print('\n'.join(map(str, af)))

Solution (Python):
from __future__ import division, print_function
from heapq import *
import sys
le = sys.__stdin__.read().split('\n')[::-1]
af = []
for ezo in range(int(le.pop())):
    (n, m, k) = list(map(int, le.pop().split()))
    ar = [set() for _ in range(n)]
    for i in range(m):
        (a, b) = list(map(int, le.pop().split()))
        ar[a - 1].add(b - 1)
        ar[b - 1].add(a - 1)
    if k * (k - 1) > 2 * m:
        af.append(-1)
    else:
        pi = [i for i in range(n) if len(ar[i]) < k]
        re = set((i for i in range(n) if len(ar[i]) >= k))
        while pi:
            a = pi.pop()
            if len(ar[a]) == k - 1:
                clique = True
                for j in ar[a]:
                    if clique:
                        for h in ar[a]:
                            if not h in ar[j] and h != j:
                                clique = False
                                break
                if clique:
                    af.append(2)
                    af.append(' '.join(map(str, [i + 1 for i in ar[a]] + [a + 1])))
                    pi = []
                    re = {-1}
            for j in ar[a]:
                ar[j].remove(a)
                if len(ar[j]) < k:
                    if j in re:
                        pi.append(j)
                        re.discard(j)
        if re and re != {-1}:
            af.append('1' + ' ' + str(len(re)))
            af.append(' '.join(map(str, [i + 1 for i in re])))
        elif re != {-1}:
            af.append(-1)
print('\n'.join(map(str, af)))