answer:下面是更正后的代码： ```c++ #include <iostream> #include <queue> #include <vector> #include <string> using namespace std; struct Node { string name; vector<Node*> neighbors; }; void bfs(Node *start) { cout << "----------Breadth First Search----------" << endl; queue<Node*> q; vector<string> visited; Node *temp = start; if (temp == NULL) return; // 检查起始节点是否为空 q.push(temp); while (!q.empty()) { Node *temp = q.front(); q.pop(); if (find(visited.begin(), visited.end(), temp->name) != visited.end()) continue; // 如果节点已被访问过则跳过 cout << temp->name << endl; visited.push_back(temp->name); for (int i=0; i<temp->neighbors.size(); i++) { if (find(visited.begin(), visited.end(), temp->neighbors[i]->name) == visited.end()) { q.push(temp->neighbors[i]); visited.push_back(temp->neighbors[i]->name); } } } } int main() { Node* nA = new Node(); nA->name = "A"; Node* nB = new Node(); nB->name = "B"; Node* nC = new Node(); nC->name = "C"; Node* nD = new Node(); nD->name = "D"; Node* nE = new Node(); nE->name = "E"; nA->neighbors.push_back(nB); nA->neighbors.push_back(nC); nB->neighbors.push_back(nD); nC->neighbors.push_back(nD); nD->neighbors.push_back(nE); bfs(nA); return 0; } ``` 更正后的代码中： 1. 包含了所需的头文件：`<iostream>`、`<queue>`、`<vector>`、`<string>`。 2. 使用 `queue<Node*>` 来保存节点，使用 `vector<string>` 来保存已访问节点的名称。 3. 添加了检查起始节点是否为空的代码（如为空则直接返回）。 4. 添加了检查节点是否已访问过的代码，以跳过已访问过的节点。 5. 在添加节点到队列时，也添加到已访问列表中，这样可以避免重复添加节点。

answer:下面是使用栈实现深度优先遍历的代码： ```c++ #include <iostream> #include <stack> #include <vector> #include <string> using namespace std; struct Node { string name; vector<Node*> neighbors; }; void dfs(Node *start) { cout << "----------Depth First Search----------" << endl; stack<Node*> s; vector<string> visited; s.push(start); while (!s.empty()) { Node *temp = s.top(); s.pop(); if (find(visited.begin(), visited.end(), temp->name) != visited.end()) continue; // 已被访问过则跳过 cout << temp->name << endl; visited.push_back(temp->name); for (int i=temp->neighbors.size()-1; i>=0; i--) { // 从右往左遍历邻居 if (find(visited.begin(), visited.end(), temp->neighbors[i]->name) == visited.end()) { s.push(temp->neighbors[i]); } } } } int main() { Node* nA = new Node(); nA->name = "A"; Node* nB = new Node(); nB->name = "B"; Node* nC = new Node(); nC->name = "C"; Node* nD = new Node(); nD->name = "D"; Node* nE = new Node(); nE->name = "E"; nA->neighbors.push_back(nB); nA->neighbors.push_back(nC); nB->neighbors.push_back(nD); nC->neighbors.push_back(nD); nD->neighbors.push_back(nE); dfs(nA); return 0; } ``` 在更正后的代码中，使用 `stack<Node*>` 来保存节点，使用 `vector<string>` 来保存已访问节点的名称。遍历节点时，从右往左遍历邻居，并在栈中添加未被访问的邻居。

answer:. Here is the SQL script to create the TransTbl table with the specified columns: ``` CREATE TABLE TransTbl( TransID INT PRIMARY KEY, CustID INT FOREIGN KEY REFERENCES CustomerTbl(CustID), CustName VARCHAR(50) FOREIGN KEY REFERENCES CustomerTbl(CustName), ProductID INT FOREIGN KEY REFERENCES ProductTbl(ProductID), ProductName VARCHAR(50) FOREIGN KEY REFERENCES ProductTbl(ProductName), ProductQTY INT FOREIGN KEY REFERENCES ProductTbl(ProductQTY), ProductMFee DECIMAL(10,2) FOREIGN KEY REFERENCES ProductTbl(ProductMFee) ); ``` Note: Please make sure that the CustomerTbl and ProductTbl tables with the necessary columns exist in your database before executing this script. Also, adjust the data types and sizes of the columns based on your specific requirements.

answer:Sure! Here's the modified SQL script with varchar(30) instead of varchar(50): ``` CREATE TABLE TransTbl( TransID INT PRIMARY KEY, CustID INT FOREIGN KEY REFERENCES CustomerTbl(CustID), CustName VARCHAR(30) FOREIGN KEY REFERENCES CustomerTbl(CustName), ProductID INT FOREIGN KEY REFERENCES ProductTbl(ProductID), ProductName VARCHAR(30) FOREIGN KEY REFERENCES ProductTbl(ProductName), ProductQTY INT FOREIGN KEY REFERENCES ProductTbl(ProductQTY), ProductMFee DECIMAL(10,2) FOREIGN KEY REFERENCES ProductTbl(ProductMFee) ); ``` Just make sure to adjust the size of the varchar columns based on your specific needs.