zoukankan      html  css  js  c++  java
  • 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;
    }
    
  • 相关阅读:
    全卷积神经网络FCN
    面试题 —— 面向对象
    【一题多解】平方根的计算及完全平方数的判断
    【一题多解】Python 字符串逆序
    JVM 自带性能监测调优工具 (jstack、jstat)及 JVM GC 调优
    n 中选 m —— 随机采样的艺术
    位图(bitmap)—— C语言实现
    C++ STL 数据结构与算法 —— 排序
    斐波那契 —— 矩阵形式推导
    常用文本编辑器 editor 的常用插件 —— CopyEdit
  • 原文地址:https://www.cnblogs.com/YDDDD/p/11620101.html
Copyright © 2011-2022 走看看