spfa  维护起点到每个点的最短路经过了几个边。

如果大于等于n肯定存在负环(否则最多n-1)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define ls (o<<1)
#define rs (o<<1|1)
#define pb push_back
const double PI= acos(-1.0);
const int M = 1e5+7;int head[M],cnt=1;
void init(){cnt=1,memset(head,0,sizeof(head));}
struct EDGE{int to,nxt,w;}ee[M*2];
void add(int x,int y,int w){ee[++cnt].nxt=head[x],ee[cnt].w=w,ee[cnt].to=y,head[x]=cnt;}int vs[M];
ll d[M];
int nm[M];
int n,m;
bool spfa()
{memset(d,0x3f,sizeof(d));memset(nm,0,sizeof(nm));memset(vs,0,sizeof(vs));queue<int>q;q.push(1);d[1]=0;vs[1]=1;while(q.size()){int x=q.front();q.pop();vs[x]=0;for(int i=head[x];i;i=ee[i].nxt){int y=ee[i].to,w=ee[i].w;if(d[y]>d[x]+w){d[y]=d[x]+w;nm[y]=nm[x]+1;//到y点的最短路，需要的路径数//	cout<<q.size()<<"   "<<vs[y]<<"  "<<"  - -   "<<d[x]<<"  "<<d[y]<<"   -  =   "<<x<<"  "<<y<<" "<< nm[y]<<endl;if(nm[y]>=n)return false; if(!vs[y])q.push(y),vs[y]=1;//	cout<<q.size()<<" =-= "<<endl;}}}return true;
}
int main()
{ios::sync_with_stdio(false);cin.tie(0);int t;cin>>t;while(t--){cin>>n>>m;init();for(int i=1;i<=m;i++){int u,v,w;cin>>u>>v>>w;if(w>=0)add(v,u,w);add(u,v,w);}if(spfa())cout<<"NO"<<endl;else cout<<"YES"<<endl;}return 0;
}