Analyzing negative SPFA ring (Wormholes)

Wormholes (SPFA negative determination ring)

Topic poj3259

Subject to the effect

Given F farm, i.e. N sets of samples. Each sample is given n, m, w are the points, the number of edges (two-way, will consume time elapsed t), the number of wormholes (unidirectional, returns after a time t).
If he can return to a certain point and time than before to reach this point is smaller, then the output "YES" Otherwise, output "NO".

Thinking

If an ordinary way, then the time counts as a positive, if a wormhole counts as negative. Then run spfa.
In spfa, if the number of times through the point (i.e. the number of enqueued point) exceeds n, then that there is a negative loop, the output yes, return. If the queue is empty, outputs NO

#include <iostream>
#include <vector>
#include <queue>

#define N 1000
#define E 3000
#define INF 0x3f3f3f
using namespace std;

vector<pair<int, int>> eg[N];
queue<int> q;
int inq[N];
int d[N];
int n, m, w;

void init()
{
    for (int i = 0; i <= n; i++)
        inq[i] = 0;
    for (int i = 0; i <= n; i++)
        eg[i].clear();
    for (int i = 0; i <= n; i++)
        d[i] = INF;
}

void spfa()
{
    q.push(0);
    inq[0]++;
    d[0] = 0;
    while (!q.empty())
    {
        int now = q.front();

        q.pop();
        for (int i = 0; i < eg[now].size(); i++)
        {
            int v = eg[now][i].first;
            if (d[v] > d[now] + eg[now][i].second)
            {
                d[v] = d[now] + eg[now][i].second;
                //cout << now << "  " << v << "   " << d[v] << endl;
                if (inq[v] > n)
                {
                    cout << "YES" << endl;
                    return;
                }
                else
                {
                    inq[v]++;
                    q.push(v);
                }
            }
        }
    }
    cout << "NO" << endl;
}

int main()
{
    int T, s, e, t;
    cin >> T;
    while (T--)
    {

        cin >> n >> m >> w;
        init();
        for (int i = 1; i <= m; i++)
        {
            cin >> s >> e >> t;
            s--, e--;
            eg[s].push_back(make_pair(e, t));
            eg[e].push_back(make_pair(s, t));
        }
        for (int i = 1; i <= w; i++)
        {
            cin >> s >> e >> t;
            s--, e--;
            eg[s].push_back(make_pair(e, -t));
        }
        spfa();
    }
}