思路：

任选一点a，和a没有边直接相连的点一定和a在同一个集合，由此构造得到一个集合A。用类似的方法再构造一个集合B，并将剩下的点放在集合C中，就得到了三个集合A，B，C。再检查A，B，C是否符合要求即可。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int N = 100005;
 4 vector<int> G[N];
 5 int res[N];
 6 int main()
 7 {
 8     int n, m;
 9     while (cin >> n >> m)
10     {
11         for (int i = 1; i <= n; i++) { res[i] = 0; G[i].clear(); }
12         int a, b;
13         for (int i = 0; i < m; i++)
14         {
15             cin >> a >> b;
16             G[a].push_back(b);
17             G[b].push_back(a);
18         }
19         res[1] = 1;
20         set<int> st{G[1].begin(), G[1].end()};
21         for (int i = 2; i <= n; i++)
22         {
23             if (!st.count(i)) res[i] = 1;
24         }
25         int i = 2;
26         for ( ; i <= n; i++)
27         {
28             if (res[i] != 1) break;
29         }
30         res[i] = 2;
31         set<int> st2{G[i].begin(), G[i].end()};
32         for (int j = 2; j <= n; j++)
33         {
34             if (j == i || res[j] == 1) continue;
35             if (st2.count(j)) res[j] = 3;
36             else res[j] = 2;
37         }
38         vector<int> c(4, 0);
39         for (int i = 1; i <= n; i++) c[res[i]]++;
40         bool flg = true;
41         if (!c[1] || !c[2] || !c[3]) flg = false;
42         if (m != c[1] * c[2] + c[2] * c[3] + c[3] * c[1]) flg = false;
43         for (int i = 1; i <= n; i++)
44         {
45             int p = res[i];
46             for (auto it: G[i])
47             {
48                 if (res[it] == p) { flg = false; break; }
49             }
50             if (!flg) break;
51         }
52         if (!flg) { cout << -1 << endl; continue; }
53         for (int i = 1; i <= n; i++) cout << res[i] << " ";
54         cout << endl;
55     }
56     return 0;
57 }