Score : $500$ points

问题描述

给定一个包含 $N$ 个顶点和 $M$ 条边的无向图 $G$。$G$ 是简单的（不含自环和多条边）且是连通的。

对于 $i = 1, 2, \ldots, M$，第 $i$ 条边是一条无向边 $\lbrace u_i, v_i \rbrace$ 连接顶点 $u_i$ 和 $v_i$。

构建两个 $G$ 的生成树 $T_1$ 和 $T_2$，满足以下两个条件。($T_1$ 和 $T_2$ 不必是不同的生成树。)

• $T_1$ 满足以下条件：

  若将 $T_1$ 视为以顶点 $1$ 为根的有根树，则对于 $G$ 中不属于 $T_1$ 的任何边 $\lbrace u, v \rbrace$，在 $T_1$ 中，$u$ 和 $v$ 中必有一个是另一个的祖先。

• $T_2$ 满足以下条件：

  若将 $T_2$ 视为以顶点 $1$ 为根的有根树，则不存在不属于 $T_2$ 的 $G$ 中的边 $\lbrace u, v \rbrace$，使得在 $T_2$ 中，$u$ 和 $v$ 中的一个是另一个的祖先。

我们可以证明总存在满足上述条件的 $T_1$ 和 $T_2$。

以上为通义千问 qwen-max 翻译，仅供参考。

Problem Statement

You are given an undirected graph $G$ with $N$ vertices and $M$ edges. $G$ is simple (it has no self-loops and multiple edges) and connected.

For $i = 1, 2, \ldots, M$, the $i$-th edge is an undirected edge $\lbrace u_i, v_i \rbrace$ connecting Vertices $u_i$ and $v_i$.

Construct two spanning trees $T_1$ and $T_2$ of $G$ that satisfy both of the two conditions below. ($T_1$ and $T_2$ do not necessarily have to be different spanning trees.)

• $T_1$ satisfies the following.

  If we regard $T_1$ as a rooted tree rooted at Vertex $1$, for any edge $\lbrace u, v \rbrace$ of $G$ not contained in $T_1$, one of $u$ and $v$ is an ancestor of the other in $T_1$.

• $T_2$ satisfies the following.

  If we regard $T_2$ as a rooted tree rooted at Vertex $1$, there is no edge $\lbrace u, v \rbrace$ of $G$ not contained in $T_2$ such that one of $u$ and $v$ is an ancestor of the other in $T_2$.

We can show that there always exists $T_1$ and $T_2$ that satisfy the conditions above.

Constraints

• $2 \leq N \leq 2 \times 10^5$
• $N-1 \leq M \leq \min\lbrace 2 \times 10^5, N(N-1)/2 \rbrace$
• $1 \leq u_i, v_i \leq N$
• All values in input are integers.
• The given graph is simple and connected.

Input

Input is given from Standard Input in the following format:

$N$ $M$

$u_1$ $v_1$

$u_2$ $v_2$

$\vdots$

$u_M$ $v_M$

Output

Print $(2N-2)$ lines to output $T_1$ and $T_2$ in the following format. Specifically,

• The $1$-st through $(N-1)$-th lines should contain the $(N-1)$ undirected edges $\lbrace x_1, y_1\rbrace, \lbrace x_2, y_2\rbrace, \ldots, \lbrace x_{N-1}, y_{N-1}\rbrace$ contained in $T_1$, one edge in each line.

• The $N$-th through $(2N-2)$-th lines should contain the $(N-1)$ undirected edges $\lbrace z_1, w_1\rbrace, \lbrace z_2, w_2\rbrace, \ldots, \lbrace z_{N-1}, w_{N-1}\rbrace$ contained in $T_2$, one edge in each line.

You may print edges in each spanning tree in any order. Also, you may print the endpoints of each edge in any order.

$x_1$ $y_1$

$x_2$ $y_2$

$\vdots$

$x_{N-1}$ $y_{N-1}$

$z_1$ $w_1$

$z_2$ $w_2$

$\vdots$

$z_{N-1}$ $w_{N-1}$

Sample Input 1

6 8
5 1
4 3
1 4
3 5
1 2
2 6
1 6
4 2

Sample Output 1

1 4
4 3
5 3
4 2
6 2
1 5
5 3
1 4
2 1
1 6

In the Sample Output above, $T_1$ is a spanning tree of $G$ containing $5$ edges $\lbrace 1, 4 \rbrace, \lbrace 4, 3 \rbrace, \lbrace 5, 3 \rbrace, \lbrace 4, 2 \rbrace, \lbrace 6, 2 \rbrace$. This $T_1$ satisfies the condition in the Problem Statement. Indeed, for each edge of $G$ not contained in $T_1$:

• for edge $\lbrace 5, 1 \rbrace$, Vertex $1$ is an ancestor of $5$;
• for edge $\lbrace 1, 2 \rbrace$, Vertex $1$ is an ancestor of $2$;
• for edge $\lbrace 1, 6 \rbrace$, Vertex $1$ is an ancestor of $6$.

$T_2$ is another spanning tree of $G$ containing $5$ edges $\lbrace 1, 5 \rbrace, \lbrace 5, 3 \rbrace, \lbrace 1, 4 \rbrace, \lbrace 2, 1 \rbrace, \lbrace 1, 6 \rbrace$. This $T_2$ satisfies the condition in the Problem Statement. Indeed, for each edge of $G$ not contained in $T_2$:

• for edge $\lbrace 4, 3 \rbrace$, Vertex $4$ is not an ancestor of Vertex $3$, and vice versa;
• for edge $\lbrace 2, 6 \rbrace$, Vertex $2$ is not an ancestor of Vertex $6$, and vice versa;
• for edge $\lbrace 4, 2 \rbrace$, Vertex $4$ is not an ancestor of Vertex $2$, and vice versa.

Sample Input 2

4 3
3 1
1 2
1 4

Sample Output 2

1 2
1 3
1 4
1 4
1 3
1 2

Tree $T$, containing $3$ edges $\lbrace 1, 2\rbrace, \lbrace 1, 3 \rbrace, \lbrace 1, 4 \rbrace$, is the only spanning tree of $G$. Since there are no edges of $G$ that are not contained in $T$, obviously this $T$ satisfies the conditions for both $T_1$ and $T_2$.

update @ 2024/3/10 10:43:48