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#include <bits/stdc++.h>
using namespace std;
const int kNmax = 50005;
const int kInf = 0x3f3f3f3f;
class Task {
public:
void solve() {
read_input();
print_output(get_result());
}
private:
int n;
int m;
int source;
vector<pair<int, int> > adj[kNmax];
void read_input() {
ifstream fin("bellmanford.in");
fin >> n >> m;
source = 1;
for (int i = 1, x, y, w; i <= m; i++) {
fin >> x >> y >> w;
adj[x].push_back(make_pair(y, w));
}
fin.close();
}
vector<int> get_result() {
/*
TODO: Gasiti distantele minime de la nodul source la celelalte noduri
folosind BellmanFord pe graful orientat cu n noduri, m arce stocat in adj.
d[node] = costul minim / lungimea minima a unui drum de la source la nodul
node;
d[source] = 0;
d[node] = -1, daca nu se poate ajunge de la source la node.
Atentie:
O muchie este tinuta ca o pereche (nod adiacent, cost muchie):
adj[x][i].first = nodul adiacent lui x,
adj[x][i].second = costul.
In cazul in care exista ciclu de cost negativ, returnati un vector gol:
return vector<int>();
*/
vector<int> d(n + 1, kInf);
vector<int> visits(n + 1, 0);
vector<bool> inQueue(n + 1, 0);
queue<int> q;
d[source] = 0;
q.push(source);
inQueue[source] = 1;
while(!q.empty()) {
int act = q.front();
q.pop();
inQueue[act] = 0;
for (auto elem: adj[act]) {
int vec = elem.first;
int cost = elem.second;
if (d[act] + cost < d[vec]) {
d[vec] = d[act] + cost;
if (inQueue[vec] == 0) {
inQueue[vec] = 1;
q.push(vec);
}
visits[vec]++;
if (visits[vec] > n) {
return vector<int>(0, n + 1);
}
}
}
}
for (int i = 1; i <= n; i++) {
if (d[i] == kInf) {
d[i] = -1;
}
}
return d;
}
void print_output(vector<int> result) {
ofstream fout("bellmanford.out");
if (result.size() == 0) {
fout << "Ciclu negativ!\n";
} else {
for (int i = 2; i <= n; i++) {
fout << result[i] << ' ';
}
fout << '\n';
}
fout.close();
}
};
// Please always keep this simple main function!
int main() {
// Allocate a Task object on heap in order to be able to
// declare huge static-allocated data structures inside the class.
Task *task = new Task();
task->solve();
delete task;
return 0;
}