You are given a directed weighted graph $G$ with $N$ vertices and $M$ edges. Find any cycle of negative weight in it, if such a cycle exists.

In another formulation of the problem you have to find all pairs of vertices which have a path of arbitrarily small weight between them.

It is convenient to use different algorithms to solve these two variations of the problem, so we'll discuss both of them here.

Bellman-Ford algorithm allows you to check whether there exists a cycle of negative weight in the graph, and if it does, find one of these cycles.

The details of the algorithm are described in the article on the Bellman-Ford algorithm. Here we'll describe only its application to this problem.

Do $N$ iterations of Bellman-Ford algorithm. If there were no changes on the last iteration, there is no cycle of negative weight in the graph. Otherwise take a vertex the distance to which has changed, and go from it via its ancestors until a cycle is found. This cycle will be the desired cycle of negative weight.

```
struct Edge {
int a, b, cost;
};
int n, m;
vector<Edge> edges;
const int INF = 1000000000;
void solve()
{
vector<int> d(n);
vector<int> p(n, -1);
int x;
for (int i = 0; i < n; ++i) {
x = -1;
for (Edge e : edges) {
if (d[e.a] + e.cost < d[e.b]) {
d[e.b] = d[e.a] + e.cost;
p[e.b] = e.a;
x = e.b;
}
}
}
if (x == -1) {
cout << "No negative cycle found.";
} else {
for (int i = 0; i < n; ++i)
x = p[x];
vector<int> cycle;
for (int v = x;; v = p[v]) {
cycle.push_back(v);
if (v == x && cycle.size() > 1)
break;
}
reverse(cycle.begin(), cycle.end());
cout << "Negative cycle: ";
for (int v : cycle)
cout << v << ' ';
cout << endl;
}
}
```

The Floyd-Warshall algorithm allows to solve the second variation of the problem - finding all pairs of vertices $(i, j)$ which don't have a shortest path between them (i.e. a path of arbitrarily small weight exists).

Again, the details can be found in the Floyd-Warshall article, and here we describe only its application.

Run Floyd-Warshall algorithm on the graph.
Initially $d[v][v] = 0$ for each $v$.
But after running the algorithm $d[v][v]$ will be smaller than $0$ if there exists a negative length path from $v$ to $v$.
We can use this to also find all pairs of vertices that don't have a shortest path between them.
We iterate over all pairs of vertices $(i, j)$ and for each pair we check whether they have a shortest path between them.
To do this try all possibilities for an intermediate vertex $t$.
$(i, j)$ doesn't have a shortest path, if one of the intermediate vertices $t$ has $d[t][t] < 0$ (i.e. $t$ is part of a cycle of negative weight), $t$ can be reached from $i$ and $j$ can be reached from $t$.
Then the path from $i$ to $j$ can have arbitrarily small weight.
We will denote this with `-INF`

.

```
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
for (int t = 0; t < n; ++t) {
if (d[i][t] < INF && d[t][t] < 0 && d[t][j] < INF)
d[i][j] = - INF;
}
}
}
```