Bellman-Ford 做法，SPFA 做法见：https://oj.33dai.cn/p/P5960/solution

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5000;
const int INF = 2147483647;
int n, m;
int M; // 边的条数
int u[MAXN * 2 + 5], v[MAXN * 2 + 5], w[MAXN * 2 + 5];
int dis[MAXN + 1 + 5];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    M = 0;
    for (int i = 1; i <= m; i++)
    {
        int op, a, b, c;
        cin >> op;
        if (op == 1)
        {
            cin >> a >> b >> c;
            // a-b>=c ~ b-a<=-c
            ++M;
            u[M] = a, v[M] = b, w[M] = -c;
        }
        if (op == 2)
        {
            cin >> a >> b >> c;
            // a-b<=c
            ++M;
            u[M] = b, v[M] = a, w[M] = c;
        }
        if (op == 3)
        {
            cin >> a >> b;
            // a=b ~ a-b<=0 , b-a<=0
            ++M;
            u[M] = b, v[M] = a, w[M] = 0;
            u[M] = a, v[M] = b, w[M] = 0;
        }
    }
    // Bellman-Ford
    for (int i = 1; i <= n; i++)
        dis[i] = 0; // 超级源点到每个点的距离都是 0
    for (int i = 1; i <= n; i++)
    {
        bool flag = true;
        for (int j = 1; j <= M; j++)
            if (dis[u[j]] + w[j] < dis[v[j]])
            {
                flag = false;
                dis[v[j]] = dis[u[j]] + w[j];
            }
        if (flag)
            break;
        if (i == n && !flag)
        {
            cout << "No";
            return 0;
        }
    }
    cout << "Yes";
    return 0;
}