B. Bear and Friendship Condition

 http://codeforces.com/problemset/problem/791/B

题意：给一个图问是否具有传递性，如1->2,2->3,有1->3则YES，否则NO

思路：dfs得到一个连通块，查看每个点的度数是否为点个数-1；

反思：st获取连通块时，st用了vector，每次取头而删除末尾=-=；

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=2e5+9;
const int mod=1e9+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

vector<int>g[maxn];
queue<int>st;
int d[maxn];
bool vis[maxn];

void dfs(int u)
{
    vis[u]=1;
    st.push(u);
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        if(!vis[v])
        {
            dfs(v);
        }
    }
}

int main()
{
    int n,m;
    cin>>n>>m;
    mem(d,0);
    for(int i=0;i<m;i++)
    {
        int x,y;
        cin>>x>>y;
        g[x].push_back(y);
        g[y].push_back(x);
        d[x]++;
        d[y]++;
    }
    mem(vis,0);
    bool f=1;
    for(int i=1;i<=n;i++)
    {
        if(!vis[i])
        {
            dfs(i);
            int ss=st.size();
            while(!st.empty())
            {
                int k=st.front();
                st.pop();
                if(d[k]!=ss-1)
                {
                    f=0;
                    break;
                }
            }
        }
        if(!f) break;
    }
    if(f) puts("YES");
    else puts("NO");
	return 0;
}