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[Source quaestionis]
https://www.acwing.com/problem/content/3590/

[Titulus descriptionis]
Dato grapho indirecto omnesque margines in eo, utrum omnes vertices huius graphi connectantur.

Input forma】
Input plures summas notitiarum continet.
Prima linea cuiusvis copiae notitiarum duos integros n et m continet, numerum punctorum et margines graphi indirectae repraesentans.
Sequuntur m lineae, quaelibet linea duos integros x, y continet, indicans punctum x et punctum y esse connexum.
Puncta numerantur ab 1 ad n.
Non potest esse in picturaduplici oreetAuto-loop。

[Output format]
Singulae notae outputs unam lineam et unum exitum.

data range】
Initus continet usque ad 10 mutatorias notitias.
1≤n≤1000;
1≤m≤5000;
1≤x,y≤n

Input sample】
4 3
1 2
2 3
3 2
3 2
1 2
2 3

Output sample】
NO
ETIAM'

Analysis Algorithmorum
The "Et collectionem quaerere"Ad codicem exsequendam singula, quaeso vide:https://blog.csdn.net/hnjzsyjyj/article/details/126455868
Principium usurae dfs ad graphum connexum determinandum in hac quaestione positum est in "dfs oportet omnia puncta percurrere graphi connexa"Si punctum non transigitur," significat non connexum.

[Algorithmus code: dfs+nexus stella deinceps]
dfs algorithmus templatehttps://blog.csdn.net/hnjzsyjyj/article/details/118736059
● For details on chain forward astra, amabo see:https://blog.csdn.net/hnjzsyjyj/article/details/139369904

#include <bits/stdc++.h>
using namespace std;
 
const int N=1e3+5;
const int M=5e3+5;
int e[M<<1],ne[M<<1],h[N],idx;
bool st[N];
int n,m;
 
void add(int a,int b) {
    e[idx]=b,ne[idx]=h[a],h[a]=idx++;
}
 
void dfs(int u) {
    st[u]=1;
    for(int i=h[u]; i!=-1; i=ne[i]) {
        int j=e[i];
        if(!st[j]) dfs(j);
    }
}
 
int main() {
    while(cin>>n>>m) {
        memset(st,false,sizeof st);
        memset(h,-1,sizeof h);
        idx=0;
        while(m--) {
            int a,b;
            cin>>a>>b;
            add(a,b),add(b,a);
        }
 
        dfs(1);
 
        bool flag=true;
        for(int i=1; i<=n; i++)
            if(!st[i]) {
                flag=false;
                break;
            }
 
        if(flag) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
 
    return 0;
}
 
 
/*
in:
4 3
1 2
2 3
3 2
3 2
1 2
2 3
out:
NO
YES
*/

[Algorithm code: dfs+adiacentia matrix]
dfs algorithmus templatehttps://blog.csdn.net/hnjzsyjyj/article/details/118736059
● Adiacentia matrix exsecutionem graphi indirectam indirectae:https://blog.csdn.net/hnjzsyjyj/article/details/116245897

#include <bits/stdc++.h>
using namespace std;
 
const int N=1010;
int g[N][N];
bool st[N];
int n,m;
 
void dfs(int u) {
    st[u]=true;
    for(int i=1; i<=n; i++)
        if(!st[i] && g[u][i]!=0) dfs(i);
}
 
int main() {
    while(cin>>n>>m) {
        memset(g,0,sizeof g);
        memset(st,false,sizeof st);
        int x,y;
        while(m--) {
            cin>>x>>y;
            g[x][y]=g[y][x]=1;
        }
        dfs(1);
        int i;
        for(i=1; i<=n; i++) {
            if(!st[i]) break;
        }
        if(i<=n) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
    return 0;
}
 
/*
in:
4 3
1 2
2 3
3 2
3 2
1 2
2 3
out:
NO
YES
*/

references】
https://www.acwing.com/solution/content/124095/
https://blog.csdn.net/hnjzsyjyj/article/details/118736059
https://blog.csdn.net/hnjzsyjyj/article/details/139369904