Bipartite graph determination

First, let's talk about what is a bipartite graph , For a picture , We divide its vertex set into two parts A and B, Thus, one of the two vertices of any edge is A One of them is in B in , Instead of having two vertices of an edge all in A In or all of B In this case .

Let's start with a theorem ： A graph is a bipartite graph if and only if there are no odd rings in this graph , That is, there is no ring with an odd number of vertices .

If we use this theorem to determine whether a graph is a bipartite graph, then we can use the method of coloring to realize , The color of the two vertices of each edge must be different , We dye the graph according to this principle , If the coloring is successful, then this graph is a bipartite graph , Otherwise, it is not a bipartite graph .

Give an example and its code implementation ：

sample input ：

4 4
1 3
1 4
2 3
2 4

sample output ：

Yes

Code ：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
using namespace std;
const int N=2e5+10;
int h[N],e[N],ne[N],color[N],idx;
//color[i] On behalf of the i Color of nodes  
void add(int x,int y)
{
	e[idx]=y;
	ne[idx]=h[x];
	h[x]=idx++;
}
bool dfs(int x,int c)
{
	color[x]=c;
	for(int i=h[x];i!=-1;i=ne[i])
	{
		int j=e[i];
		if(color[j]!=-1)//j Nodes have been dyed  
		{
			if(color[j]==c) return false;// Two vertices of an edge have the same color , contradiction  
		}
		else if(!dfs(j,c^1)) return false;// Find contradictions  
	}
	return true;// All nodes return true without finding contradictions  
}
int main()
{
	int n,m;
	cin>>n>>m;
	memset(h,-1,sizeof h);
	int u,v;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		add(u,v);add(v,u);
	}
	memset(color,-1,sizeof color);// The initial color is -1 
	bool flag=true;
	for(int i=1;i<=n;i++)
	{
		if(color[i]==-1)
		{
			if(!dfs(i,1))// Dye the current dot into 1 Number color  
			{
				flag=false;// Exit directly when there is an odd ring  
				break;
			}
		}
	}
	if(flag) puts("Yes");
	else puts("No");
	return 0;
}