#include <bits/stdc++.h>using namespace std;const int N = 5005;int ideg[N], odeg[N], n, m, edg[N][N];
bool visited[N];void travel(int u, int distance)
{cout << u << " " << distance << endl; visited[u] = true;for (int v = 1; v <= n; v++)if (edg[u][v] != -1 && !visited[v])//是否已经访问过 travel(v, distance + edg[u][v]); //if there is an edge (u, v) and v has not been visited, then travel(v)
}
int main()
{cin >> n >> m;memset(edg, -1, sizeof edg);memset(visited, false, sizeof visited);for (int u, v, w, i = 1; i <= m; i++)cin >> u >> v >> w, edg[u][v] = w, odeg[u]++, ideg[v]++;//出度和入度 for (int i = 1; i <= n; i++)cout << ideg[i] << " " << odeg[i] << endl;for (int i = 1; i <= n; i++)if (!visited[i]) travel(i, 0);
}/*
Given a graph with N nodes and M unidirectional edges.
Each edge e_i starts from u_i to v_i and weights w_i
Output a travelsal from node 1 and output degree of each node.
*/

#include <bits/stdc++.h>using namespace std;const int N = 5005;struct edge {int u, v, w; edge *next;//next指针指向 edge(int _u, int _v, int _w, edge *_next):u(_u), v(_v), w(_w), next(_next) {}
};
edge *head[N]; //List[u] 最前面的节点是谁
int ideg[N], odeg[N], n, m;
bool visited[N];void add(int u, int v, int w)
{edge *e = new edge(u, v, w, head[u]);head[u] = e;
}
void travel(int u, int distance)
{cout << u << " " << distance << endl; visited[u] = true;for (edge *e = head[u]; e ; e = e -> next)if (!visited[e -> v])travel(e -> v, distance + e -> w); //if there is an edge (u, v) and v has not been visited, then travel(v)
}
int main()
{cin >> n >> m;memset(visited, false, sizeof visited);memset(head, 0, sizeof head);for (int u, v, w, i = 1; i <= m; i++)cin >> u >> v >> w, add(u, v, w), odeg[u]++, ideg[v]++;for (int i = 1; i <= n; i++)cout << ideg[i] << " " << odeg[i] << endl;for (int i = 1; i <= n; i++)if (!visited[i]) travel(i, 0);
}/*
Given a graph with N nodes and M unidirectional edges.
Each edge e_i starts from u_i to v_i and weights w_i
Output a travelsal from node 1 and output degree of each node.
*/

#include <bits/stdc++.h>using namespace std;const int N = 5005;struct edge {int u, v, w, next;
}edg[N];
int head[N]; //List[u] stores all edges start from u
int ideg[N], odeg[N], n, m, cnt; //cnt: numbers of edges
bool visited[N];void add(int u, int v, int w)
{int e = ++cnt;edg[e] = (edge){u, v, w, head[u]};head[u] = e;
}
void travel(int u, int distance)
{cout << u << " " << distance << endl; visited[u] = true;for (int e = head[u]; e ; e = edg[e].next)if (!visited[edg[e].v])travel(edg[e].v, distance + edg[e].w); //if there is an edge (u, v) and v has not been visited, then travel(v)
}
int main()
{cin >> n >> m; cnt = 0;memset(visited, false, sizeof visited);memset(head, 0, sizeof head);for (int u, v, w, i = 1; i <= m; i++)cin >> u >> v >> w, add(u, v, w), odeg[u]++, ideg[v]++;for (int i = 1; i <= n; i++)cout << ideg[i] << " " << odeg[i] << endl;for (int i = 1; i <= n; i++)if (!visited[i]) travel(i, 0);
}/*
Given a graph with N nodes and M unidirectional edges.
Each edge e_i starts from u_i to v_i and weights w_i
Output a travelsal from node 1 and output degree of each node.
*/

#include <bits/stdc++.h>using namespace std;const int N = 5005;struct edge {int u, v, w;
};
vector<edge> edg[N]; //edge记录变长数组记录的是什么类型
int ideg[N], odeg[N], n, m, cnt; //cnt: numbers of edges
bool visited[N];void add(int u, int v, int w)
{edg[u].push_back((edge){u, v, w});//一个强制类型转换
}
void travel(int u, int distance)
{cout << u << " " << distance << endl; visited[u] = true;for (int e = 0; e < edg[u].size(); e++)//遍历边 if (!visited[edg[u][e].v])//以u出发的第e条出边 travel(edg[u][e].v, distance + edg[u][e].w); //if there is an edge (u, v) and v has not been visited, then travel(v)
}
int main()
{cin >> n >> m; cnt = 0;memset(visited, false, sizeof visited);for (int u, v, w, i = 1; i <= m; i++)cin >> u >> v >> w, add(u, v, w), odeg[u]++, ideg[v]++;for (int i = 1; i <= n; i++)cout << ideg[i] << " " << odeg[i] << endl;for (int i = 1; i <= n; i++)if (!visited[i]) travel(i, 0);
}/*
Given a graph with N nodes and M unidirectional edges.
Each edge e_i starts from u_i to v_i and weights w_i
Output a travelsal from node 1 and output degree of each node.
*/