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.
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.
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).
4 3
1 3
3 4
1 4
YES
4 4
3 1
2 3
3 4
1 2
NO
10 4
4 3
5 10
8 9
1 2
YES
3 2
1 2
2 3
NO
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.
题目链接：http://codeforces.com/contest/791/problem/B
分析：给你m对朋友关系; 如果x-y是朋友,y-z是朋友，要求x-z也是朋友. 问你所给的图是否符合！
用DFS去找他们之间的关系，有向图去算m个点对应的边的数量，看是否相等！
下面给出AC代码：
1 #include<bits/stdc++.h> 2 typedef long long ll; 3 using namespace std; 4 const int N=150000+100; 5 int a[N]; 6 int b[N]; 7 int vis[N]; 8 vector<int>vc[N];//定义一个向量 9 ll t; 10 void DFS(int x) 11 { 12 t++;//计算有关的节点没有被标记过 13 vis[x]=1; 14 for(int j=0;j<vc[x].size();j++) 15 { 16 if(vis[vc[x][j]]==0)//如果和它相连的点没有被标记， 17 DFS(vc[x][j]); 18 } 19 } 20 int main() 21 { 22 int m,n; 23 int x,y; 24 scanf("%d%d",&n,&m); 25 memset(vis,0,sizeof(vis)); 26 memset(b,0,sizeof(b)); 27 for(int i=0;i<m;i++) 28 { 29 scanf("%d%d",&x,&y); 30 vc[x].push_back(y);//找到x和y的关系 31 vc[y].push_back(x); 32 b[x]=1; 33 b[y]=1; 34 35 } 36 ll sum=0; 37 for(int i=1;i<=n;i++) 38 { 39 if(vis[i]==0&&b[i]) 40 { 41 t=0; 42 DFS(i); 43 sum=sum+t*(t-1);//(t-1)*t求边数，完全图 44 } 45 } 46 if(2*m!=sum)cout<<"NO"<<endl;//N个完全图的边数等于m个节点所构成的边数 47 else 48 cout<<"YES"<<endl; 49 }
