zoukankan      html  css  js  c++  java
  • CodeForce-791B Bear and Friendship Condition(并查集)

    Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).

    There are n members, numbered 1 through n. m pairs of members are friends. Of course, a member can't be a friend with themselves.

    Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.

    For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.

    Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.

    Input

    The first line of the input contain two integers n and m (3 ≤ n ≤ 150 000, ) — the number of members and the number of pairs of members that are friends.

    The i-th of the next m lines contains two distinct integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). Members ai and bi are friends with each other. No pair of members will appear more than once in the input.

    Output

    If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).

    Example
    Input
    4 3
    1 3
    3 4
    1 4
    Output
    YES
    Input
    4 4
    3 1
    2 3
    3 4
    1 2
    Output
    NO
    Input
    10 4
    4 3
    5 10
    8 9
    1 2
    Output
    YES
    Input
    3 2
    1 2
    2 3
    Output
    NO
    Note

    The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not。

    题意:给定N个点M条链路来描述两个点之间的关系,并且A-B,B-C,那么A-C一定要有边,问你给定的符不符合要求

    思路:并查集,把连在一起的统计到一棵树上,然后树上所有点的边数都应该相等

    #include <iostream>
    #include <cstdio>
    using namespace std;
    int fa[100050], du[100050], ran[100050];
    int find(int a)
    {
        return fa[a] == a ? a : find(fa[a]);
    }
    void bing(int x, int y)
    {
        x = find(x);
        y = find(y);
        if (x != y)
        {
            fa[x] = y;
        }
    }
    int main()
    {
        int n, m;
        cin >> n >> m;
        for (int i = 0; i <= n; i++)
        {
            fa[i] = i, du[i] = 0, ran[i] = 0;
        }
        for (int i = 0; i < m; i++)
        {
            int a, b;
            cin >> a >> b;
            bing(a, b);
            du[a]++, du[b]++;
        }
        for (int i = 0; i <= n; i++)
        {
            int x = find(i);
            ran[x]++;
        }
        int flag = 0;
        for (int i = 0; i <= n; i++)
        {
            int x = find(i);
            if (du[i] != ran[x] - 1)
            {
                flag = 1;
                break;
            }
        }
        if (!flag)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
        return 0;
    }

     

  • 相关阅读:
    【翻译自mos文章】 11gR1版本号 asmcmd的新命令--cp、md_backup、md_restore
    Android实现ListView或GridView首行/尾行距离屏幕边缘距离
    iOS-为方便项目开发在pch加入一些经常使用宏定义
    [ACM] FZU 1686 神龙的难题 (DLX 反复覆盖)
    Cocos2d-x Touch事件处理机制
    在linux環境下安裝jprofiler_linux_8_0_2.sh
    QT5 Failed to load platform plugin &quot;windows&quot; 终极解决方式 命令行问题
    我们想要如何子的生活?
    javaEE mvc样例具体解释
    安装Kali Linux操作系统Kali Linux无线网络渗透
  • 原文地址:https://www.cnblogs.com/YingZhixin/p/6648215.html
Copyright © 2011-2022 走看看