CF1228D Complete Tripartite

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：

 #include <bits/stdc++.h>
 using namespace std;
 const int N = ;
 vector<int> G[N];
 int res[N];
 int main()
 {
     int n, m;
     while (cin >> n >> m)
     {
         for (int i = ; i <= n; i++) { res[i] = ; G[i].clear(); }
         int a, b;
         for (int i = ; i < m; i++)
         {
             cin >> a >> b;
             G[a].push_back(b);
             G[b].push_back(a);
         }
         res[] = ;
         set<int> st{G[].begin(), G[].end()};
         for (int i = ; i <= n; i++)
         {
             if (!st.count(i)) res[i] = ;
         }
         int i = ;
         for ( ; i <= n; i++)
         {
             if (res[i] != ) break;
         }
         res[i] = ;
         set<int> st2{G[i].begin(), G[i].end()};
         for (int j = ; j <= n; j++)
         {
             if (j == i || res[j] == ) continue;
             if (st2.count(j)) res[j] = ;
             else res[j] = ;
         }
         vector<int> c(, );
         for (int i = ; i <= n; i++) c[res[i]]++;
         bool flg = true;
         if (!c[] || !c[] || !c[]) flg = false;
         if (m != c[] * c[] + c[] * c[] + c[] * c[]) flg = false;
         for (int i = ; i <= n; i++)
         {
             int p = res[i];
             for (auto it: G[i])
             {
                 if (res[it] == p) { flg = false; break; }
             }
             if (!flg) break;
         }
         if (!flg) { cout << - << endl; continue; }
         for (int i = ; i <= n; i++) cout << res[i] << " ";
         cout << endl;
     }
     return ;
 }