CodeForces - 771A Bear and Friendship Condition （并查集）

最新推荐文章于 2020-06-16 16:07:13 发布

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最新推荐文章于 2020-06-16 16:07:13 发布

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分类专栏：思维题／模拟／乱搞／贪心并查集

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博客介绍了如何使用并查集解决CodeForces上的771A问题，该问题涉及判断一个社交网络是否满足合理性的条件——如果X和Y是朋友，Y和Z也是朋友，那么X和Z也应该是朋友。通过检查网络中是否存在不满足此条件的三元组，来确定答案是'YES'还是'NO'。博客内容包括问题描述、输入输出格式以及可能的解决方案策略。

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题目链接：http://codeforces.com/problemset/problem/771/A点击打开链接

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, $\text{[math]}$ ) — 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).

   Examples
 

     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.

$\text{[math]}$

判断是不是满足各个形成的关系图是不是完全图而完全图的边数是n*（n-1）

开一个数组记录每个点链接的个数依次判断每个点将所有点的度数根据关系式计算看是否能够形成完全图与k相等就可以

注意这里不用判断0阶完全图跟1阶完全图

#include <bits/stdc++.h>
using namespace std;
long long int pre[200000];
long long int check[200000];
int findx(int x)
{
    int r=x;
    while(pre[r]!=r)
    {
        r=pre[r];
    }

    int i=x;int j;
    while(pre[i]!=r)
    {
        j=pre[i];
        pre[i]=r;
        i=j;
    }

    return r;
}
void join(int x,int y)
{
    int p1=findx(x);
    int p2=findx(y);
    if(p1!=p2)
    {
        pre[p2]=p1;
    }
}
int main()
{
   int n,k;
   scanf("%d%d",&n,&k);
   for(int i=0;i<=n;i++)
        pre[i]=i;
   for(int ii=0;ii<k;ii++)
   {
       int m,mm;
       scanf("%d%d",&m,&mm);
       join(m,mm);
   }
   for(int i=1;i<=n;i++)
   {
        check[findx(i)]++;
   }
   long long int sum=0;
   for(int i=1;i<=n;i++)
   {
       if(check[i]!=0&&check[i]!=1)
            sum+=(check[i]*(check[i]-1)/2);
   }
   if(sum==k)
     cout << "YES";
   else
    cout <<"NO";
}