怎么修改这个算法double GetWeight(ALGraph G, int nStart, int nEnd, char cKind) { double dTorM = INFINITY*1.0; if (cKind == 'T' || cKind == 'M') { if (cKind == 'T') { ArcNode* pT = G.vertices[nStart].Firstarc; for (; pT != NULL; pT = pT->nextArc) { if (pT->nAdjvex == nEnd) { if (pT->Information.dTime < dTorM) dTorM = pT->Information.dTime; } } } else{ ArcNode* pM = G.vertices[nStart].Firstarc; for (; pM != NULL; pM = pM->nextArc) { if (pM->nAdjvex == nEnd) { if (pM->Information.dCost < dTorM) dTorM = pM->Information.dCost; } } } return dTorM; } else { cout << "Input error! Please enter T or M." << endl; cout << " If Time please enter T. If Money please enter M." << endl; cin >> cKind; return GetWeight(G, nStart, nEnd, cKind); } }

6-1算法DepthFirstSearch1

int index ; // 顶点序号，从 0 开始编号 Edge *first ; // 指向第一条邻接边的指针 } ; 在计算机科学中，图是一种数据结构，用于表示对象之间的关系。邻接矩阵是存储图的一种方法，它通过二维数组来表示图中各个...

C++实现Dijkstra算法

这个C++实现的Dijkstra算法可以有效地解决从特定起点到图中其他所有节点的最短路径问题，适用于各种实际场景，如网络路由、地图导航等。注意，由于算法依赖优先队列（通常用堆实现）来快速找到最小距离节点，实际...

如何修改double GetWeight(ALGraph G, int nStart, int nEnd, char cKind) { else { cout << "error,please input again" << endl; cout << " If Time please enter T. If Money please enter M." << endl; cin >> cKind; double dTorM = GetWeight(G, nStart, nEnd, cKind); return dTorM; } }

double GetWeight(ALGraph G, int nStart, int nEnd, char cKind) { double dTorM = 0.0; if (cKind == 'T' || cKind == 'M') { // 根据用户输入的类型，获取对应边的权值 if (cKind == 'T') { dTorM = G.arcs...

import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); int sum = 0; Express[] express = new Express[n]; for (int i = 0; i < n; i++) { String name = scanner.next(); if (name == "SL") { express[i] = new SLExpress(scanner.nextInt()); sum += express[i].getTotal(); System.out.println(sum); } else if(name == "DD") { express[i] = new DDExpress(scanner.nextInt()); sum += express[i].getTotal(); } } System.out.print(sum); } } abstract class Express { private int weight; Express(int weight) { this.weight = weight; } public int getWeight() { return this.weight; } public abstract int getTotal(); } class SLExpress extends Express { SLExpress(int weight) { super(weight); } public int getTotal() { int total = 0; if(super.getWeight() <= 1) total = 12; else total = 12 + (super.getWeight() - 1)2; return total; } } class DDExpress extends Express { DDExpress(int weight) { super(weight); } public int getTotal() { int total = 0; if(super.getWeight() <= 1) total = 5; else total = 5 + (super.getWeight() - 1)1; return total; } }

这段代码实现了一个简单的快递计价程序，其中用到了抽象类和多态的概念。程序中定义了一个Express抽象类，其中包含一个weight属性表示快递重量和一个getTotal()抽象方法用于计算运费。SLExpress和DDExpress类都继承...

解释以下这段代码并说出他的计算复杂度import java.util.*; public class 1806 { static int n; static int[] t = new int[10]; static int[] telegraph = new int[50005]; static int[] dis = new int[50005]; static int[] pre = new int[50005]; static boolean[] vis = new boolean[50005]; static ArrayList<Integer> path = new ArrayList<Integer>(); public static void main(String[] args) { Scanner sc = new Scanner(System.in); n = sc.nextInt(); for (int i = 0; i < 10; i++) { t[i] = sc.nextInt(); } for (int i = 1; i <= n; i++) { telegraph[i] = sc.nextInt(); } dijkstra(1); if (dis[n] == Integer.MAX_VALUE) { System.out.println("-1"); } else { System.out.println(dis[n]); getPath(n); System.out.println(path.size()); for (int i = 0; i < path.size(); i++) { System.out.print(path.get(i) + " "); } } } private static void dijkstra(int s) { Arrays.fill(dis, Integer.MAX_VALUE); dis[s] = 0; for (int i = 1; i <= n; i++) { pre[i] = i; } for (int k = 0; k < n; k++) { int u = -1; int minDis = Integer.MAX_VALUE; for (int i = 1; i <= n; i++) { if (!vis[i] && dis[i] < minDis) { u = i; minDis = dis[i]; } } if (u == -1) { break; } vis[u] = true; for (int i = 1; i <= n; i++) { if (!vis[i]) { int w = getWeight(telegraph[u], telegraph[i]); if (dis[u] + w < dis[i]) { dis[i] = dis[u] + w; pre[i] = u; } } } } } private static int getWeight(int a, int b) { int weight = 0; String s1 = String.valueOf(a); String s2 = String.valueOf(b); int len = Math.min(s1.length(), s2.length()); for (int i = 0; i < len; i++) { if (s1.charAt(i) != s2.charAt(i)) { weight = t[i]; break; } } return weight; } private static void getPath(int u) { if (u != 1) { getPath(pre[u]); } path.add(u); } }

这段代码实现了一个基于Dijkstra算法的最短路径问题。其中，telegraph数组存储了n个节点的电报号码，t数组存储了电报号码每一位的开销，dis数组存储了每个节点到源节点的最短距离，pre数组存储了每个节点的前驱节点...

import java.util.*; public class 1806 { static int n; static int[] t = new int[10]; static int[] telegraph = new int[50005]; static int[] dis = new int[50005]; static int[] pre = new int[50005]; static boolean[] vis = new boolean[50005]; static ArrayList<Integer> path = new ArrayList<Integer>(); public static void main(String[] args) { Scanner sc = new Scanner(System.in); n = sc.nextInt(); for (int i = 0; i < 10; i++) { t[i] = sc.nextInt(); } for (int i = 1; i <= n; i++) { telegraph[i] = sc.nextInt(); } dijkstra(1); if (dis[n] == Integer.MAX_VALUE) { System.out.println("-1"); } else { System.out.println(dis[n]); getPath(n); System.out.println(path.size()); for (int i = 0; i < path.size(); i++) { System.out.print(path.get(i) + " "); } } } private static void dijkstra(int s) { Arrays.fill(dis, Integer.MAX_VALUE); dis[s] = 0; for (int i = 1; i <= n; i++) { pre[i] = i; } for (int k = 0; k < n; k++) { int u = -1; int minDis = Integer.MAX_VALUE; for (int i = 1; i <= n; i++) { if (!vis[i] && dis[i] < minDis) { u = i; minDis = dis[i]; } } if (u == -1) { break; } vis[u] = true; for (int i = 1; i <= n; i++) { if (!vis[i]) { int w = getWeight(telegraph[u], telegraph[i]); if (dis[u] + w < dis[i]) { dis[i] = dis[u] + w; pre[i] = u; } } } } } private static int getWeight(int a, int b) { int weight = 0; String s1 = String.valueOf(a); String s2 = String.valueOf(b); int len = Math.min(s1.length(), s2.length()); for (int i = 0; i < len; i++) { if (s1.charAt(i) != s2.charAt(i)) { weight = t[i]; break; } } return weight; } private static void getPath(int u) { if (u != 1) { getPath(pre[u]); } path.add(u); } }解释一下该代码的运行过程

这段代码实现了一个基于Dijkstra算法的最短路径问题求解，输入的数据包括电报码的位数n，每个数字所需的时间t[i]，以及n个电报码。程序首先调用dijkstra函数，计算从起点1到终点n的最短路径，同时记录每个节点的前驱...

#include<iostream> #include<string> using namespace std; class Pet { protected: string name; int age; double weight; public: //Pet(string name,int age,float weight) // { // this->name=name; //this->age=age; //this->weight=weight; // } Pet() { name=""; age=0; weight=0.0; } Pet(const string&petName,int petAge,double petWeight) { name = petName; age = petAge; weight = petWeight; } string getName() { return name; } int getAge() { return age; } double getWeight() { return weight; } void setName(string name) { this->name=name; } void setAge(int age) { this->age=age; } void setWeight(double weight) { this->weight=weight; } virtual string getLifespan() { return "unknown lifespan"; } }; class Dog:public Pet { private: string breed; public: Dog(string name,int age,double weight,string breed) : Pet(name,age,weight) { this->breed=breed; } string getBreed() { return breed; } void setBreed(string breed) { this->breed=breed; } virtual string getLifespan() { if (weight>100) { return "Approximately 7 years"; } else { return "Approximately 13 years"; } } }; class Rock:public Pet { public: Rock(string name,int age,double weight) : Pet(name,age,weight) {} virtual string getLifespan() { return"Thousands of years"; } }; int main() { Petpet1=new Rock("Peter",1000,1000); Petpet2=new Dog("Mike",3,80,"Golden Retriever"); Pet*pet3=new Dog("Ham",5,80,"Labrador Retriever"); cout<getName()<<", "<getLifespan()<< endl; pet1->setAge(2); pet1->setWeight(1.0); cout<getName()<<"," <getLifespan()<<endl; cout<<"Breed: "getBreed()<<endl; pet2->setWeight(85.0); cout<getName()<<"," <getLifespan()<<endl; cout<<"Breed: "<getBreed() <<endl; pet3->setWeight(110.0); return 0; } 这段代码无法运行

这段代码其实是可以运行的，但是存在一些小问题。首先，代码开头的 #include 需要在每一行的末尾添加一个分号，例如 #include;。其次，Pet 类中的构造函数不应该被注释掉，因为在 Rock 和 Dog 类中都需要...

public class Rft { private Map<Integer, List<Edge>> adjacencyList; public Rft() { this.adjacencyList = new HashMap<>(); } public void addEdge(int u, int v, double w) { Edge edge = new Edge(v, w); if (!adjacencyList.containsKey(u)) { adjacencyList.put(u, new ArrayList<>()); } adjacencyList.get(u).add(edge); } public List<Integer> getNeighbors(int node) { List<Integer> neighbors = new ArrayList<>(); if (adjacencyList.containsKey(node)) { for (Edge edge : adjacencyList.get(node)) { neighbors.add(edge.to); } } return neighbors; } public BigDecimal getWeight(int u, int v) { BigDecimal weight = BigDecimal.valueOf(0.0); if (adjacencyList.containsKey(u)) { for (Edge edge : adjacencyList.get(u)) { if (edge.to == v) { weight = BigDecimal.valueOf(edge.weight); break; } } } return weight; } private static class Edge { int to; double weight; public Edge(int to, double weight) { this.to = to; this.weight = weight; } }解释这段代码

这段代码实现了一个有向加权图的表示，其中包含了Rft类和内部静态类Edge。 Rft类有一个私有成员变量adjacencyList，它是一个Map，用于存储每个节点的邻接列表，其中每个节点对应一个键值，对应的值是一个由Edge对象...

这个循环为什么无法终止while (cKind != 'T' && cKind != 'M') { cout << "Input error! Please enter T or M." << endl; cout << "If Time please enter T. If Money please enter M." << endl; cin >> cKind; }

double GetWeight(ALGraph G, int nStart, int nEnd, char cKind) { double dTorM = numeric_limits<double>::infinity(); while (cKind != 'T' && cKind != 'M') { cout ! Please enter T or M." ; cout << "If...

#include <iostream> #include <vector> #include <algorithm> class Animal { public: Animal(std::string name, double height, double weight, int age) : m_name(name), m_height(height), m_weight(weight), m_age(age) {} std::string getName() const { return m_name; } double getHeight() const { return m_height; } double getWeight() const { return m_weight; } int getAge() const { return m_age; } private: std::string m_name; double m_height; double m_weight; int m_age; }; bool sortByWeight(const Animal &a, const Animal &b) { return a.getWeight() < b.getWeight(); } int main() { std::vector<Animal> animals; animals.push_back(Animal("Elephant", 3.0, 5000.0, 10)); animals.push_back(Animal("Lion", 1.5, 200.0, 5)); animals.push_back(Animal("Giraffe", 5.0, 800.0, 8)); animals.push_back(Animal("Hippopotamus", 1.8, 1500.0, 12)); std::sort(animals.begin(), animals.end(), sortByWeight); for (const auto &animal : animals) { std::cout << animal.getName() << " weighs " << animal.getWeight() << " kg." << std::endl; } return 0; }能给我讲讲这段代码吗？

这段代码定义了一个 Animal 类，它表示一个动物对象，包含了动物的名字、身高、体重和年龄等属性。然后，定义了 sortByWeight 函数，它是一个自定义的比较器，用来按照动物的重量进行排序。在主函数中，我们定义了...

abstract class Express{ public int weight; public abstract int getWeight(); public abstract int getTotal(); } public class SLExpress extends Express{ public int cost; public SLExpress(){ } public SLExpress(int weight){ super.weight=weight; } public int getTotal(int weight){ cost=12+(weight-1)*2; return cost; } } public class DDExpress extends Express{ public int cost; public DDExpress(){ } public DDExpress(int weight){ super.weight=weight; } public int getTotal(int weight){ cost=5+weight-1; return cost; } } public class CaicaiStation{ public int sum; public int calculate(Express[] ex){ for(int i=0;i<ex.length;i++){ sum+=ex[i]; } } }纠错

在这段代码中，CaicaiStation类中的calculate()方法中有一个错误： 1. 在sum+=ex[i];语句中，ex[i]是Express类型的对象，不能直接进行加法操作，需要调用ex[i].getTotal()方法来获取该快递的总费用。 ...

请精修这段代码#include <iostream>#include <vector>#include <algorithm>using namespace std;struct item { int weight; int value;};bool cmp(item i1, item i2) { return i1.value * i2.weight > i2.value * i1.weight;}pair<int, vector<int>> knapsack(vector<item>& items, int capacity) { int n = items.size(); vector<int> selected_items(n, 0); int total_value = 0; sort(items.begin(), items.end(), cmp); for (int i = 0; i < n; i++) { if (capacity == 0) break; if (items[i].weight <= capacity) { selected_items[i] = 1; total_value += items[i].value; capacity -= items[i].weight; } else { selected_items[i] = capacity / items[i].weight; total_value += selected_items[i] * items[i].value; capacity -= selected_items[i] * items[i].weight; } } return make_pair(total_value, selected_items);}int main() { vector<item> items = {{2, 10}, {3, 15}, {5, 20}}; int capacity = 8; auto result = knapsack(items, capacity); cout << "Total value: " << result.first << endl; cout << "Selected items: "; for (int i : result.second) { cout << i << " "; } cout << endl; return 0;}

这段代码已经很精简了，但还是可以做一些微调，以下是我建议的改进： 1. 将头文件的顺序改为先算法库再是标准库，遵循头文件依赖关系原则。 2. 结构体 item 可以使用 class 替代，同时将结构体内部的成员变量声明...

int main() { vector<Animal> vec; vec.push_back(new Animal("tiger", 1.2, 250, 5)); vec.push_back(new Animal("lion", 1.0, 200, 6)); vec.push_back(new Animal("elephant", 2.5, 5000, 25)); vec.push_back(new Animal("giraffe", 5.5, 800, 12)); sort(vec.begin(), vec.end(), compareWeight); for (vector<Animal>::iterator iter = vec.begin(); iter != vec.end(); ++iter) { cout << (iter)->getName() << ": " << (iter)->getWeight() << endl; } cout << "The heaviest animal: " << vec.back()->getName() << endl; for (vector<Animal>::iterator iter = vec.begin(); iter != vec.end(); ++iter) { delete iter; } return 0; } 这段代码是什么意思？

这段代码是一个简单的 C++ 程序，它演示了如何使用 STL 中的 vector 和 sort 函数对一组 Animal 对象进行排序并输出结果。程序中定义了一个 Animal 类，每个 Animal 对象都有名称、体重、价格和年龄等属性。程序创建...

public int getWeight() { return weight; }

这段代码是Food类中的一个公有方法getWeight，该方法返回一个整型的weight属性值。通过定义这个方法，可以让其他类访问Food对象的weight属性，而不需要将weight属性设置为public类型。这样可以保护weight属性，...

#include<iostream> #include<fstream> using namespace std; class Dog { public: int getAge(){return a;} int getWeight(){return w;} void setdog(int x,int y){a=x;w=y;} private: int a,w; }; int main() { int v1,v2; Dog dog1; dog1.setdog(10,5); ofstream outFile("outfile.txt",ios::out); outFile<<dog1.getAge()<<" "<<dog1.getWeight(); outFile.close(); Dog dog2; ifstream inFile("outfile.txt",ios::in); inFile>>v1;inFile.seekg(1,ios::cur);inFile>>v2; dog2.setdog(v1,v2); outFile.close(); cout<<"dog1's age:"<<dog1.getAge()<<endl; cout<<"dog1's weight:"<<dog1.getWeight()<<endl; cout<<"dog2's age:"<<dog2.getAge()<<endl; cout<<"dog2's weight:"<<dog1.getWeight()<<endl; }

这是一个使用文件流进行读写的 C++ 程序。程序中定义了一个 Dog 类，包含两个私有成员变量 a 和 w，以及三个公有成员函数 getAge、getWeight 和 setdog。其中，getAge 和 getWeight 用于获取私有成员变量 a 和 w 的...

续写代码public abstract class Animal { private String name; public Animal(String name) { this.name = name; } public String getName() { return name; } public void setName(String name) { this.name = name; } public abstract void makeSound(); } // 包 animal package animal; public class Cat extends Animal { private int age; public Cat(String name, int age) { super(name); this.age = age; } public int getAge() { return age; } public void setAge(int age) { this.age = age; } @Override public void makeSound() { System.out.println("Meow"); } } public class Dog extends Animal { private String breed; public Dog(String name, String breed) { super(name); this.breed = breed; } public String getBreed() { return breed; } public void setBreed(String breed) { this.breed = breed; } @Override public void makeSound() { System.out.println("Woof"); } } public class Elephant extends Animal { private int weight; public Elephant(String name, int weight) { super(name); this.weight = weight; } public int getWeight() { return weight; } public void setWeight(int weight) { this.weight = weight; } @Override要求0.必须有抽象父类，且被继承 1.必须包含 >=2 个包2.必须创建 >=5 个类(其中包含抽象父类)3.类中属性皆为私有属性，需定义读写方法 4.子类需要重写个别父类方法

续写代码： // 包 animal package animal; public abstract class Animal ... public int getWeight() { return weight; } public void setWeight(int weight) { this.weight = weight; } @Override ＊＊＊

实验内容及要求：实现一个仿真农场，农场包括猫、狗、猪、马等动物。下面给出一个哺乳动物的类，要求编写包括猫、狗、猪、马等动物的的类，并实现以下要求：其中狗属哺乳动物，且它的属性有品种之分(在哺乳类基础上增加品种数属性)，叫声区别于其他动物(输出“Woof!”)。令猫、马、猪也属于哺乳动物，其叫声分别为:“Meow!”，“Winnie!”，“Oink!”。编程分别使各个动物表现为不一样的行为。定义一个接口 Pet，猫和狗类需要实现 Pet 的所有方法。要求 Pet 包含如会摇尾巴(增加方法，输出 “Tail wagging…”)，乞讨食物(增加方法，输出“begging for food…”)等宠物行为。哺乳动物类如下所示: public class Mammal{ protected int age = 2; protected int weight = 5; public Mammal(){ System.out.println(“Mammal constructor”); } int getAge(){ return age; } void setAge(int a){ age = a; } int getWeight(){ Acts(); class i class Nose(); extends i() 5 class Of76(); interface i(x) 7 class Clowns(); implements i[x] 7 public class Picasso(); Acts i.iMethod(x) public int iMethod(); Nose i(x).iMethod[] public int iMethod{} Of76 i[x].iMethod() public int iMethod(){ Clowns i[x].iMethod[] public int iMethod(){} Picasso Of76[] i = new Nose[3]; Of76[3] i; Nose[] i = new Nose(); Nose[] i = new Nose[3]; return weight; } void setWeight(int w){ weight = w; } void move(){ System.out.println(“Mammal move one step”); } void speak(){ System.out.println(“Mammal speak!”); } void sleep(){ System.out.println(“shhh, I’m sleeping.”); }

if(animal instanceof Pet){ Pet pet = (Pet)animal; pet.TailWagging(); pet.BeggingForFood(); } System.out.println(); } } } 输出结果应该如下： Mammal constructor Cat constructor Mammal ...

命令手册 Linux常用命令

【超强组合】基于VMD-雪融优化算法SAO-Transformer-GRU的光伏预测算研究Matlab实现.rar

1.版本：matlab2014/2019a/2024a 2.附赠案例数据可直接运行matlab程序。 3.代码特点：参数化编程、参数可方便更改、代码编程思路清晰、注释明细。 4.适用对象：计算机，电子信息工程、数学等专业的大学生课程设计、期末大作业和毕业设计。替换数据可以直接使用，注释清楚，适合新手

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这个循环为什么无法终止while (cKind != 'T' && cKind != 'M') { cout << "Input error! Please enter T or M." << endl; cout << "If Time please enter T. If Money please enter M." << endl; cin >> cKind; }

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