[数据结构与算法]西农大 数据结构 习题

题目描述
患者到医院看病的顺序是:先排队等候,再看病治疗。要求设计一个算法,模拟病人等候就诊的过程。
其中:“病人到达”用命令“?A?”(或“?a?")表示,“护士让下一位就诊”用“?N?”(或"?n?")表示,“不再接收病人排队”用“?S?"(或“?s?")表示。
输入描述:
?A?(或?a?)?病历号
?N?(或?n?)?s?(或?s?)
?x?(其它字符)
输出描述:
病历号为×的病人就诊
若队列中有数据,输出队首元素,否则输出"无病人就诊”不再接收病人排队,并输出当前队列中所有病历号
输入命令不合法!

?

?

#include <iostream>
using namespace std;

template <class T>
struct LinkNode
{
	T data;
	LinkNode* link;
	LinkNode() {
		link = NULL;
	}
	LinkNode(const T& item) {
		data = item;
		link = NULL;
	}
};

template <class T>
class Queue {
private:
	LinkNode<T>* const first = new LinkNode<T>();
	int length;
public:
	Queue() {
		length = 0;
	}

	bool Insert(T& a) {
		LinkNode<T>* newnode = new LinkNode<T>(a);
		LinkNode<T>* current = first;
		for (int i = 1; i <= length; i++) {
			current = current->link;
		}
		current->link = newnode;
		length++;
		cout << "病历号为" << current->link->data << "的患者就诊" << endl;
		return true;
	}

	bool Insert2(T& a) {
		LinkNode<T>* newnode = new LinkNode<T>(a);
		LinkNode<T>* current = first;
		for (int i = 1; i <= length; i++) {
			current = current->link;
		}
		current->link = newnode;
		length++;
		//cout << "病历号为" << current->link->data << "的患者就诊" << endl;
		return true;
	}

	bool Remove() {
		LinkNode<T>* current = first;
		if (current->link == NULL) {
			cout << "无病人就诊" << endl;
		}
		else {
			current = current->link;
			first->link = current->link;
			delete current;
			length--;
		}
		return true;
	}

	void Print() {
		LinkNode<T>* current = first;
		cout << "今天不再接收病人排队，下列排队的病人依次就诊：";
		for (int i = 1; i <= length; i++) {
			current = current->link;
			cout << current->data << " ";
		}
		cout << endl;
	}
};

int main() {
	char ch;
	int num;
	Queue<int> q1;
	cin >> ch;
	while (1) {
		switch (ch)
		{
		case 'A':
			cin >> num;
			q1.Insert(num);
			break;
		case 'a':
			cin >> num;
			q1.Insert2(num);
			break;
		case 'N':
		case 'n':
			q1.Remove();
			break;
		case 'S':
		case 's':
			q1.Print();
			return 0;
			break;
		default:
			cout << "输入命令不合法！" << endl;
			cin >> ch;
			continue;
			break;
		}
		cin >> ch;
	}
}

样例太少，无法通过oj