目前是得了40分，判负环的思路：用一个cnt[n]来存储从1到这个点n要走的步数，如果走到某一个点的步数比点的数量减一（n-1）还要大的话，就说明存在负环，否则步数最大为n-1

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 2010, M = 3010*2;
int h[N], e[M], ne[M], w[M], idx;
int times, n, m;
int dist[N], cnt[N];
bool st[N];

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

bool spfa()
{
    queue <int> q;
    q.push(1);
    st[1] = true;

    while (q.size())
    {
        int t = q.front();
        q.pop();
        st[t] = false;

        for (int i = h[t]; i != -1; 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 true;
                if (!st[j])
                {
                    q.push(j);
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main()
{
    cin >> times;

    while (times--)
    {
        idx = 0;
        memset(h, -1, sizeof h);
        memset(e, 0, sizeof e);
        memset(ne, 0, sizeof ne);
        memset(cnt, 0, sizeof cnt);
        memset(st, false, sizeof st);
        memset(dist,0,sizeof dist);
        cin >> n >> m;
        while (m--)
        {
            int a, b, c;
            cin >> a >> b >> c;
            add(a, b, c);
            if (c >= 0) add(b, a, c);
        }

        if (spfa()) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}