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.
题意：是否对于所有的子图都是完全图
题解：只需检查是否所有的点都和其他所有的点有连边即可
设子图中点的个数为cnt,只需判断所有点入度之和e是否满足cnt*(cnt-1)==e即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 int gi() 7 { 8 int str=0;char ch=getchar(); 9 while(ch>'9'||ch<'0')ch=getchar(); 10 while(ch>='0'&&ch<='9')str=str*10+ch-48,ch=getchar(); 11 return str; 12 } 13 const int N=150005; 14 int num=0; 15 int head[N];int du[N];bool mark[N]; 16 struct Lin 17 { 18 int next,to; 19 }a[N*2]; 20 21 void init(int x,int y) 22 { 23 a[++num].next=head[x]; 24 a[num].to=y; 25 head[x]=num; 26 } 27 int n,m;int cnt=0,e=0; 28 29 void dfs(int x) 30 { 31 if(mark[x])return ; 32 mark[x]=true; 33 cnt++;e+=du[x]; 34 int u; 35 for(int i=head[x];i;i=a[i].next) 36 { 37 u=a[i].to; 38 dfs(u); 39 } 40 } 41 bool work() 42 { 43 for(int i=1;i<=n;i++) 44 { 45 if(mark[i])continue; 46 cnt=0;e=0; 47 dfs(i); 48 if((long long)cnt*(cnt-1)!=e)return false; 49 } 50 return true; 51 } 52 int main() 53 { 54 n=gi();m=gi(); 55 int x,y; 56 for(int i=1;i<=m;i++) 57 { 58 x=gi();y=gi(); 59 init(x,y);init(y,x); 60 du[x]++;du[y]++; 61 } 62 if(work())printf("YES"); 63 else printf("NO"); 64 return 0; 65 }