Let $G=G(V,E)$ be a simple graph with vertex set $V$ and edge set $E$. Suppose $|V|=n$. A map $f:\,V\rightarrow\mathbb{Z}$ is called good, if $f$ satisfies the followings:
(1) $\sum_{v\in V} f(v)=|E|$;
(2) color arbitarily some vertices into red, one can always find a red vertex $v$ such that $f(v)$ is no more than the number of uncolored vertices adjacent to $v$.
Let $m(G)$ be the number of good maps. Prove that if every vertex in $G$ is adjacent to at least one another vertex, then $n\leq m(G)\leq n!$.
1. **Understanding the Problem:**
   We are given a simple graph \( G = G(V, E) \) with vertex set \( V \) and edge set \( E \). The number of vertices is \( |V| = n \). A map \( f: V \rightarrow \mathbb{Z} \) is called good if it satisfies:
   - \(\sum_{v \in V} f(v) = |E|\)
   - For any arbitrary coloring of some vertices into red, there exists a red vertex \( v \) such that \( f(v) \) is no more than the number of uncolored vertices adjacent to \( v \).

   We need to prove that if every vertex in \( G \) is adjacent to at least one other vertex, then \( n \leq m(G) \leq n! \), where \( m(G) \) is the number of good maps.

2. **Condition (2) Analysis:**
   Suppose we have a good mapping \( f \) of graph \( G \). Color all vertices red. By condition (2), there must be a vertex \( v_0 \) such that \( f(v_0) \leq 0 \). Now uncolor \( v_0 \). Again, by condition (2), there must be some vertex \( v_1 \) such that \( f(v_1) \) is no more than the number of uncolored adjacent vertices to \( v_1 \), which is at most 1. Hence, \( f(v_1) \leq 1 \). Continuing in this manner, we get a sequence of vertices \( v_0, v_1, \ldots, v_n \) such that \( f(v_i) \leq \min\{i, d(v_i)\} \), where \( d(v_i) \) denotes the degree of vertex \( v_i \).

3. **Inductive Proof for \( n \leq m(G) \):**
   - **Base Case:** For \( |V| = 2 \), there is nothing to prove.
   - **Inductive Step:** Assume it is true for \( |V| = k \) and consider \( |V| = k+1 \). Take any vertex \( v \) and remove it and its \( d(v) \) edges. By our assumption, there are at least \( n-1 \) good mappings of the new graph. For any of these good mappings, when we reattach \( v \), consider it as \( v_{k+1} \) in the vertex sequence, and it must satisfy property (2) by the previous analysis. Furthermore, we have \( |E| - |E'| = d(v_{k+1}) \), so let \( f(v_{k+1}) = d(v_{k+1}) \) and we also satisfy condition (1). This gives \( n-1 \) mappings of \( G \).

   We selected \( v \) arbitrarily, so we can do the same for all vertices. Moreover, when we select a vertex \( v \), we generate \( n-1 \) mappings where \( f(v) = d(v) \). So suppose we take a vertex \( v \) for one of the mappings \( f_1 \), there must be a vertex \( u \) such that \( f_1(u) < d(u) \) (from (1)). Then take vertex \( u \) and generate another \( n-1 \) mappings \( f_2, f_3, \ldots, f_n \). These must be distinct from \( f_1 \) because \( f_k(u) = d(u) \) for \( k > 1 \). So the inductive step is complete.

4. **Upper Bound \( m(G) \leq n! \):**
   - **Nice Maps:** Call a map *nice* if it satisfies condition (2), and instead of condition (1), we have \(\sum_{v \in V} f(v) = N \geq |E|\). Suppose we have a nice map \( f \) for which \(\sum_{v \in V} f(v) = N > |E|\). If we take any vertex \( v \) and reduce \( f(v) \) by 1, then we preserve property (2) and \( N \) falls to \( N-1 \geq |E| \). On the other hand, if we have a vertex \( v \) and increase \( f(v) \) by 1, we do not necessarily preserve condition (2). It follows that the number of nice maps for a fixed \( N \) cannot increase as \( N \) increases.

   - **Inductive Proof:** It's trivial for \( |V| = 2 \). Assume true for \( |V| = k \) and consider \( G \) with \( |V| = k+1 \). Take some good mapping of \( G \). From the previous analysis, we can select some vertex \( v_{k+1} \) and remove it and preserve condition (2). Furthermore, \( f(v_{k+1}) \) is uniquely determined by the other values of the function by condition (1). Then we must have \(\sum_{i=0}^k f(v_k) = N \geq |E| - d(v_{k+1}) \geq |E| - n\).

   So we have that the map on \( G \setminus v_{k+1} \) is nice. When \( N = |E| - d(v_{k+1}) \), the map is good (in the original sense), so by our assumption, there are no more than \( (n-1)! \) such maps. As \( N \) increases to \( |E| - 1 \), at each fixed \( N \), we can have no more than \( (n-1)! \) maps. Consequently, \( G \setminus v_{k+1} \) can have no more than \( n \cdot (n-1)! = n! \) nice maps, implying \( G \) has no more than \( n! \) good maps.

The final answer is \( \boxed{ n \leq m(G) \leq n! } \)