Ideas

One ： When a ring with odd edges is a graph, it is not a bipartite graph

Two ： Dye all dots in two different colors , If it is a bipartite graph, the two colors must be evenly distributed

3、 ... and ： Color conflict in the dyeing process is not a bipartite graph

Four ： The connected next node of each node has the opposite color to that node

5、 ... and ： If a node is not stained , Just dye him and all his connections

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N=10e5+10,M=2*N;

int h[N],e[M],ne[M],idx;
int st[N];
int n,m;

void add(int a,int b)
{
    e[idx]=b;
    ne[idx]=h[a];
    h[a]=idx++;
}

bool dfs(int x,int c)
{
    st[x]=c;
    for(int i=h[x];i!=-1;i=ne[i])
    {
        int t=e[i];
        if(!st[t])  {
            if(!dfs(t,3-c)) return false ;
        }
        else if(st[t]==c)return false ;
    }
    return true;
}


int main()
{
    cin>>n>>m;
    memset(h, -1, sizeof h);
    while (m -- ){
        
        int a,b;
        cin>>a>>b;
        add(a, b);
        add(b,a);
    }
    bool flag =true;
    for(int i=1;i<=n;i++)
    {
        if(!st[i])
        {
            if(!dfs(i,1))
            {
                flag=false;
                break;
            }
        }
    }
    if(flag) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    
    return 0;
}