codeforces 1228D - Complete Tripartite

// three points chart, quite simply, speaking at ideas: first of all the points are divided into three collections, 
under // judge the collection can not be null point in the set can not have a connection, and finally judge the lower number 
# the include <bits./stdc ++ H.> 
the using namespace STD; 
typedef Long Long I64; 
const int = 1E5 + MAXN 32; 
int n-, m, U, V, COL [MAXN]; 
SET <int> Grape [MAXN]; 
I64 ARR [. 8]; 
int main () 
{ 
    iOS :: sync_with_stdio (to false); cin.tie (0), cout.tie (0); 
    CIN >> >> n-m; 
    for (int I = 0; I =! m; I ++) 
    { 
        CIN >> >> U V; 
        Grape [U] .insert (V); 
        Grape [V] .insert (U); 
    } 
    COL [. 1] =. 1; 
    for (int I = 2; I <= n-; I ++) 
        IF (! Grape [. 1] .count (I)) 
            COL [I] =. 1; 
    for (int I = 2; I <= n-;++i)
        If the node i if // not accessed (COL [i]!) 
        { 
            COL [i] = 2; // to find a set of 2 
            for (int. 1 = J; J <= n-; J ++) 
            { 
                IF (! Grape [I] .count (J)) 
                { 
                    IF (COL [J] ==. 1) 
                    { 
                        COUT << - <<. 1 '\ n-'; 
                        return 0; 
                    } 
                    COL [J] = 2; 
                } 
            } 
            BREAK; 
        } // the second mark 
    for (int I =. 1; I <= n-; I ++) 
        IF (COL [I]!) 
            COL [I] =. 3; 
    for (int I =. 1; I <= n-; ++ I) 
    {
        for(auto v :Grape[i])
            if(col[v] == col[i])
            {
                cout<<-1<<'\n';
                return 0;
            }
    }//集合内的边去重
    for(int i=1;i<=n;++i)
        ++arr[col[i]];
    if(!arr[1]||!arr[2]||!arr[3])
        cout<<"-1"<<'\n';
    else if((arr[1]*arr[2]+arr[1]*arr[3]+arr[2]*arr[3])!=m)
        cout<<-1<<"\n";
    else{
        for(int i=1;i<=n;++i)
            cout<<col[i]<<" ";
        cout<<'\n';
    }
}