思路：建立超级源点，设置为0，将它到任何点的距离都设置为0，则可以发现，若无负环并且最小边<=0时，可以求出最小距离，但若最小边大于0时，提前输出最小边，
看这组数据：3 3
1 2 3
2 3 2
1 3 1
若按照刚刚的判，输出是0，程序默认把最小距离当成设置的超级源点到任何一个点的距离了，至于如何判负环，定义lop数组来存深度，若深度大于n，则有边被访问两次，有负环。
#include <bits/stdc++.h>
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define ll long long
#define ld long double
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define MP make_pair
#define PI pair<int,int>
using namespace std;
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
int t, h[N], cnt, dis[N], vis[N], lop[N], n, m;
struct node {
int v, w, net;
} no[N];
void add(int u, int v, int w) {
no[cnt].v = v;
no[cnt].w = w;
no[cnt].net = h[u];
h[u] = cnt++;
}
int spfa() {
queue<int>q;
q.push(0);
vis[0] = 1, dis[0] = 0, lop[0] = 0;
while(!q.empty()) {
int u = q.front();
q.pop();
vis[u] = 0;
for(int i = h[u]; i != -1; i = no[i].net) {
int v = no[i].v, w = no[i].w;
if(dis[v] > dis[u] + w) {
dis[v] = dis[u] + w;
lop[v] = lop[u] + 1;
if(lop[v] > n)
return 1;
if(!vis[v]) {
q.push(v);
vis[v] = 1;
}
}
}
}
return 0;
}
int main() {
cin >> t;
while(t--) {
ll ans = 0x3f3f3f3f3f3f3f3f;
mem(h, -1), mem(dis, 0x3f), me(lop), me(vis), cnt = 0;
cin >> n >> m;
for(int i = 1; i <= n; i++)
add(0, i, 0);
for(int u, v, w, i = 0; i < m; i++) {
cin >> u >> v >> w;
add(u, v, w);
ans = min(ans, (ll)w);
}
if(ans >= 0) {
cout << ans << endl;
continue;
}
if(spfa())
cout << "-inf" << endl;
else {
for(int i = 1; i <= n; i++)
ans = min(ans, (ll)dis[i]);
cout << ans << endl;
}
}
return 0;
}