无向图有四个结点分别是1，2，3，4，其中1和2，4连通，2和1，3，4连通，3和2，4连通，4和，1，2，3连通1、实现如下功能：已知无向图的邻接表，判断该邻接表是否连通。 //1. 提示：函数可以写成如下形式BOOL judge_is_connected(adjList b[ ]) //2. 提示：（a）结构体自己定义；（b）定义提示1的函数

以下是实现该功能的代码： ```c++ #include <iostream> #include <vector> using namespace std; const int MAXN = 100; struct node { int v; node *next; }; struct adjList { int data; node *first; }; bool visited[MAXN]; void DFS(adjList b[], int s) { visited[s] = true; node *p = b[s].first; while (p != nullptr) { int v = p->v; if (!visited[v]) { DFS(b, v); } p = p->next; } } bool judge_is_connected(adjList b[]) { int n = sizeof(b) / sizeof(b[0]); for (int i = 1; i <= n; i++) { visited[i] = false; } DFS(b, 1); for (int i = 1; i <= n; i++) { if (!visited[i]) { return false; } } return true; } int main() { adjList b[5]; for (int i = 1; i <= 4; i++) { b[i].data = i; b[i].first = nullptr; } node *p1 = new node; p1->v = 2; p1->next = nullptr; b[1].first = p1; node *p2 = new node; p2->v = 1; p2->next = nullptr; node *p3 = new node; p3->v = 3; p3->next = nullptr; node *p4 = new node; p4->v = 4; p4->next = nullptr; b[2].first = p2; b[2].first->next = p3; b[2].first->next->next = p4; node *p5 = new node; p5->v = 2; p5->next = nullptr; node *p6 = new node; p6->v = 4; p6->next = nullptr; b[3].first = p5; b[3].first->next = p6; node *p7 = new node; p7->v = 1; p7->next = nullptr; node *p8 = new node; p8->v = 2; p8->next = nullptr; node *p9 = new node; p9->v = 3; p9->next = nullptr; b[4].first = p7; b[4].first->next = p8; b[4].first->next->next = p9; if (judge_is_connected(b)) { cout << "The adjList is connected." << endl; } else { cout << "The adjList is not connected." << endl; } return 0; } ``` 其中，DFS 函数表示深度优先遍历，judge_is_connected 函数判断邻接表是否连通。在主函数中，先初始化邻接表，然后调用 judge_is_connected 函数判断邻接表是否连通。