#include <cstdio>
#include <cstring>
#define N 1010
#define M 2010
#define INF 0x3f3f3f3f
int d[N],u[M],v[M],w[M];
int n,m;int BF(int s)  //Bellman-Ford
{int k,i,j;for(i=0; i<n; i++) d[i]=INF;d[s]=0;for(k=1; k<n; k++)  //进行V-1次松弛for(int i=0; i<m; i++)  //枚举所有的边
        {int x=u[i],y=v[i];if( d[x]+w[i] < d[y] )d[y]=d[x]+w[i];}int OK=1;for(int i=0; i<m; i++)  //检查一遍所有的边
    {int x=u[i],y=v[i];if( d[x]+w[i] < d[y] )  //还能进行松弛
        {OK=0;break;}}return OK;
}
int main()
{int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(int i=0; i<m; i++)scanf("%d%d%d",&u[i],&v[i],&w[i]);if(!BF(0))  //有负环printf("possible\n");elseprintf("not possible\n");}return 0;
}

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define N 1010
#define NN 5
#define M 2010
#define INF 0x3f3f3f3f
int n,m;
int d[N],f[N],nnext[M],u[M],v[M],w[M];  //用数组来模拟邻接表，并且使用头插法
int c[N];  //记录每个顶点进队的次数
bool vis[N];  //标记顶点在队内void input()
{scanf("%d%d",&n,&m);memset(f,0,sizeof(f));//memset(nnext,0,sizeof(nnext));for(int i=1; i<=m; i++)  //边集数组从下标1保存到m
    {scanf("%d%d%d",&u[i],&v[i],&w[i]);nnext[i]=f[u[i]];  //头插法f[u[i]]=i;        //头插法
    }
/*printf("打印邻接表:\n");for(int i=0; i<n; i++){printf("%d:\n",i);int j=f[i];while(j!=0){printf("%d %d\n",v[j],w[j]);j=nnext[j];}}
*/return ;
}int spfa_bfs(int s)
{queue <int> q;memset(d,0x3f,sizeof(d));d[s]=0;memset(c,0,sizeof(c));memset(vis,0,sizeof(vis));q.push(s);  vis[s]=1; c[s]=1;//顶点入队vis要做标记，另外要统计顶点的入队次数int OK=1;while(!q.empty()){int x;x=q.front(); q.pop();  vis[x]=0;//队头元素出队，并且消除标记for(int k=f[x]; k!=0; k=nnext[k]) //遍历顶点x的邻接表
        {int y=v[k];if( d[x]+w[k] < d[y]){d[y]=d[x]+w[k];  //松弛if(!vis[y])  //顶点y不在队内
                {vis[y]=1;    //标记c[y]++;      //统计次数q.push(y);   //入队if(c[y]>NN)  //超过入队次数上限，说明有负环return OK=0;}}}}return OK;}
int main()
{int T;scanf("%d",&T);while(T--){input();if(!spfa_bfs(0))  //有负环printf("possible\n");elseprintf("not possible\n");}return 0;
}

//时间最快0.008
#include <cstdio>
#include <cstring>
#define N 1010
#define M 2010
#define INF 0x3f3f3f3f
int d[N],f[N];
bool vis[N];
struct edge
{int u,v,w,next;}e[M];
int n,m;void input()
{scanf("%d%d",&n,&m);memset(f,0,sizeof(f));for(int i=1; i<=m; i++)  //读入所有边
    {scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);int u=e[i].u;e[i].next=f[u];f[u]=i;}
/*printf("打印邻接表：\n");for(int i=0; i<n; i++){printf("%d:\n",i);for(int k=f[i]; k!=0; k=e[k].next)printf("%d %d\n",e[i].v,e[i].w);}
*/return ;
}int spfa_dfs(int u)
{vis[u]=1;for(int k=f[u]; k!=0; k=e[k].next){int v=e[k].v,w=e[k].w;if( d[u]+w < d[v] ){d[v]=d[u]+w;if(!vis[v]){if(spfa_dfs(v))return 1;}elsereturn 1;}}vis[u]=0;return 0;
}
int main()
{int T;scanf("%d",&T);while(T--){input();for(int i=0; i<n; i++){ vis[i]=0; d[i]=INF;}d[0]=0;if(spfa_dfs(0))  printf("possible\n");else     printf("not possible\n");}return 0;
}    

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
#define N 1010
#define NN 5
#define M 2010
#define INF 0x3f3f3f3f
int n,m;
int d[N],f[N],nnext[M],u[M],v[M],w[M];  //用数组来模拟邻接表，并且使用头插法
int c[N];  //记录每个顶点进队的次数
bool vis[N];  //标记顶点在队内void input()
{scanf("%d%d",&n,&m);memset(f,0,sizeof(f));//memset(nnext,0,sizeof(nnext));for(int i=1; i<=m; i++)  //边集数组从下标1保存到m
    {scanf("%d%d%d",&u[i],&v[i],&w[i]);nnext[i]=f[u[i]];  //头插法f[u[i]]=i;        //头插法
    }
/*printf("打印邻接表:\n");for(int i=0; i<n; i++){printf("%d:\n",i);int j=f[i];while(j!=0){printf("%d %d\n",v[j],w[j]);j=nnext[j];}}
*/return ;
}int spfa_bfs(int s)
{queue <int> q;memset(d,0x3f,sizeof(d));d[s]=0;memset(c,0,sizeof(c));memset(vis,0,sizeof(vis));q.push(s);  vis[s]=1; c[s]=1;//顶点入队vis要做标记，另外要统计顶点的入队次数int OK=1;while(!q.empty()){int x;x=q.front(); q.pop();  vis[x]=0;//队头元素出队，并且消除标记for(int k=f[x]; k!=0; k=nnext[k]) //遍历顶点x的邻接表
        {int y=v[k];if( d[x]+w[k] < d[y]){d[y]=d[x]+w[k];  //松弛if(!vis[y])  //顶点y不在队内
                {vis[y]=1;    //标记c[y]++;      //统计次数q.push(y);   //入队if(c[y]>NN)  //超过入队次数上限，说明有负环return OK=0;}}}}return OK;}
int main()
{int T;scanf("%d",&T);while(T--){input();int s;for(s=0; s<n; s++)  //枚举所有源点
        {int tmp=spfa_bfs(s) ;//for(int i=0; i<n; i++)//printf("%d ",d[i]);//printf("\n");if(!tmp)  break;}if(s<n)  //有负环printf("possible\n");elseprintf("not possible\n");}return 0;
}