O(nm)

int h[N], w[M], e[M], ne[M], idx;
int dist[N], cnt[N];
bool s[N];

void add(int a, int b, int c) //memset(h,-1,sizeof h);
{
    e[idx] = b;
    ne[idx] = h[a];
    w[idx] = c;
    h[a] = idx++;
}

bool spfa() //返回1时有负环
{
    memset(dist, 0, sizeof dist);
    memset(cnt, 0, sizeof cnt);
    memset(s, 0, sizeof s);
    queue<int> q;
    for (int i = 1; i <= n; i++)
    {
        q.push(i);
        s[i] = 1;
    }
    while (q.size())
    {
        int t = q.front();
        q.pop();
        s[t] = 0;
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (dist[j] > dist[t] + w[i])
            {
                dist[j] = dist[t] + w[i];
                cnt[j] = cnt[t] + 1;
                if (cnt[j] >= n)
                {
                    return 1;
                }
                if (!s[j])
                {
                    q.push(j);
                    s[j] = 1;
                }
            }
        }
    }
    return 0;
}