Codeforces Round #589 (Div. 2) D. Complete Tripartite(染色)

链接:

https://codeforces.com/contest/1228/problem/D

题意:

You have a simple undirected graph consisting of n vertices and m edges. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.

Let's make a definition.

Let v1 and v2 be two some nonempty subsets of vertices that do not intersect. Let f(v1,v2) be true if and only if all the conditions are satisfied:

There are no edges with both endpoints in vertex set v1.
There are no edges with both endpoints in vertex set v2.
For every two vertices x and y such that x is in v1 and y is in v2, there is an edge between x and y.
Create three vertex sets (v1, v2, v3) which satisfy the conditions below;

All vertex sets should not be empty.
Each vertex should be assigned to only one vertex set.
f(v1,v2), f(v2,v3), f(v3,v1) are all true.
Is it possible to create such three vertex sets? If it's possible, print matching vertex set for each vertex.

思路:

先染色, 然后根据度数判断是否满足每个都联通.

代码:

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5+10;

vector<int> G[MAXN];
int col[MAXN], deg[MAXN];
int n, m;

int main()
{
    cin >> n >> m;
    int u, v;
    for (int i = 1;i <= m;i++)
    {
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
        deg[u]++;
        deg[v]++;
    }
    for (int i = 1;i <= n;i++)
        col[i] = 1;
    for (int i = 1;i <= n;i++)
    {
        if (col[i] == 1)
        {
            for (int j = 0; j < G[i].size(); ++j)
            {
                if (col[G[i][j]] == 1)
                    col[G[i][j]] = 2;
            }
        }
    }
    for (int i = 1;i <= n;i++)
    {
        if (col[i] == 2)
        {
            for (int j = 0; j < G[i].size(); ++j)
            {
                if (col[G[i][j]] == 2)
                    col[G[i][j]] = 3;
            }
        }
    }
    vector<int> Num[4];
    for (int i = 1;i <= n;i++)
        Num[col[i]].push_back(i);
    if (Num[3].size() == 0 || Num[2].size() == 0 || Num[1].size() == 0)
    {
        puts("-1");
        return 0;
    }
    bool flag = true;
    for (int i = 1;i <= n;i++)
    {
        int sum = 0;
        for (int j = 1;j <= 3;j++)
        {
            if (j == col[i])
                continue;
            sum += Num[j].size();
        }
        if (deg[i] != sum)
        {
            flag = false;
            break;
        }
    }
    if (!flag)
        puts("-1");
    else
    {
        for (int i = 1;i <= n;i++)
            cout << col[i] << ' ';
    }
    puts("");

    return 0;
}