这样似乎跑得快：
初始化所有的dis是0，然后枚举每个点作为起点，用DFS更新所有点的dis；
如果更新到一个栈中节点，那么有负环。
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f3f
#define space putchar(' ')
#define enter putchar('
')
using namespace std;
typedef long long ll;
template <class T>
bool read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
else if(c == EOF) return 0;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
return 1;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}
const int N = 200005;
int T, n, m;
int ecnt, adj[N], go[2*N], nxt[2*N];
ll w[2*N], dis[N];
bool ins[N], done;
void add(int u, int v, int ww){
go[++ecnt] = v;
nxt[ecnt] = adj[u];
adj[u] = ecnt;
w[ecnt] = ww;
}
void dfs(int u){
if(done) return;
ins[u] = 1;
for(int e = adj[u], v; e; e = nxt[e]){
if(done) return;
if(v = go[e], w[e] < dis[v] - dis[u]){
dis[v] = dis[u] + w[e];
if(ins[v]) return (void)(done = 1);
else dfs(v);
}
}
ins[u] = 0;
}
void init(){
for(int i = 1; i <= n; i++)
ins[i] = adj[i] = dis[i] = 0;
ecnt = done = 0;
}
int main(){
read(T);
while(T--){
read(n), read(m);
init();
for(int i = 1, u, v, ww; i <= m; i++){
read(u), read(v), read(ww);
add(u, v, ww);
if(ww >= 0) add(v, u, ww);
}
for(int i = 1; i <= n; i++){
dfs(i);
if(done) break;
}
if(done) puts("YE5");
else puts("N0");
}
return 0;
}