讓標題：農民希望貓左岸用船、狗、魚運至右岸。

在運輸過程中，每次只能運送一個單純的動物。過河農民可以空船。

當中當人不在此岸時，狗會咬貓。貓會吃魚。當人在此岸時，則不會發生沖突。

請用面向對象的設計思想解決此類問題。

分析：通過題設條件能夠得出下面結論：1、左到右，不存在從左岸到右岸的空船（也就是從左岸到右岸必須運送一僅僅動物）；2、右到左。有且僅僅有兩種情況：①空船過河、②帶一僅僅動物過河。

程序設計：5個類：MyCrossRiver.java、CrossProcess.java、CrossStep.java、Status.java、Animal.java。當中。MyCrossRiver是程序執行的主類，也包括了過河邏輯；CrossProsess是記錄整個過程已經走過的正確的步驟序列；CrossStep表示的是所走過的每一步；Status表示的是對當前步驟的狀態的封裝（對于左岸和右岸的數據是深度拷貝）；Animal是動物的抽象。

主要代碼例如以下：

MyCrossRiver.java：

package com.others;import java.util.ArrayList; import java.util.List;/*** 貓狗魚過河問題* @author xuefeihu**/ public class MyCrossRiver {/** 河左岸 **/private List<Animal> left = new ArrayList<Animal>();/** 河右岸 **/private List<Animal> right = new ArrayList<Animal>();/** 人的位置：左邊是true,右邊是false **/private boolean flag = true;/** 過河步驟 **/private CrossProcess process = new CrossProcess();public static void main(String[] args) {new MyCrossRiver().crossRiver();}/*** 初始化條件*/public void initAnimal(){Animal dog = new Animal("狗", 1, -1, 2);Animal cat = new Animal("貓", 2, 1, 3);Animal fish = new Animal("魚", 3, 2, -1);left.add(dog);left.add(cat);left.add(fish);}/*** 過河操作*/public void crossRiver(){initAnimal();while(right.size() != 3){Status preStatus = new Status(this.left, this.right, this.flag);//記錄步驟前狀態CrossStep step = new CrossStep(process.getStepCount()+1, this.flag, null, preStatus);//創建步驟if(this.flag){//從左到右過河（不存在空船過河）int leftIndex = step.getNextLeftIndex();int leftSize = this.left.size();if(leftIndex >= leftSize){//回退數據this.process.removeLastStep();CrossStep step2 = this.process.getLastStep();this.back2Step(step2);continue;}else{//帶動物過河step.setAnimal(this.left.get(leftIndex));}}else{//從右往左過河Animal animal = null;boolean rightSecurity = this.check(right);if(rightSecurity){animal = this.getTargetAnimal(this.right);if(animal == null){//沖突無法解決時，回退數據this.process.removeLastStep();CrossStep step2 = this.process.getLastStep();this.back2Step(step2);continue;}else{step.setAnimal(animal);}}else{//無沖突時。不運送動物step.setAnimal(null);}}boolean result = moveByOneStep(step);if(!result){//假設運行失敗，則恢復上一步驟this.process.removeLastStep();CrossStep step2 = this.process.getLastStep();this.back2Step(step2);}}this.process.printStepMessage();}/*** 移動操作* @param step* @return 返回true表示轉移成功，false表示轉移失敗（失敗時須要回退到上一步進行轉移）*/public boolean moveByOneStep(CrossStep step){/** 返回的結果：true表示移動成功、false表示失敗 **/boolean result = false;/** 循環標志位 **/boolean cricleFlag = false;while(!cricleFlag){int leftIndex = step.getNextLeftIndex();int leftSize = step.getStatus().getLeft().size();int rightIndex = step.getNextRightIndex();if(this.flag){//當能夠找到下一個索引時，進行轉移if(leftIndex < leftSize){//帶動物過河Animal animal = left.remove(leftIndex);right.add(animal);flag = !flag;step.setAnimal(animal);}else if(leftIndex >= leftSize){return false;//返回失敗信息，并交給上一層程序處理}}else if(!this.flag){if(step.getAnimal() == null){//此時能夠單人過河flag = !flag;this.process.addStep(step);return true;}else{//帶動物過河Animal animal = right.remove(rightIndex);left.add(animal);flag = !flag;}}//檢查沖突情況（轉移后回退）if(!this.flag && check(this.left)){step.addNextLeftIndex();this.back2Step(step);}else if(this.flag && check(this.right)){step.addNextRightIndex();this.back2Step(step);}else {this.process.addStep(step);result = true;cricleFlag = true;}}return result;}/*** 將當前狀態恢復到step步驟* @param step*/private void back2Step(CrossStep step){Status status = step.getStatus();this.left = status.getLeft();this.right = status.getRight();this.flag = status.getFlag();}/*** 從沖突的數據中獲取不沖突的動物：不存在時返回null*/public Animal getTargetAnimal(List<Animal> array){Animal result = null;//克隆對象List<Animal> lists = new ArrayList<Animal>();Animal target = null;Animal source = null;for(int i = 0; i < array.size(); i++){source = array.get(i);target = new Animal(source.type, source.id, source.afraid, source.control);lists.add(target);}//查找對象for(int i = 0; i < lists.size(); i++){result = lists.remove(i);if(!check(lists)){break;}lists.add(i, result);}return result;}/*** 檢查是否有沖突*/private boolean check(List<Animal> array){boolean result = true;if(array.size() > 1){for(int i = 0; i < array.size(); i++){for(int j = i+1; j < array.size(); j++){result = array.get(i).check(array.get(j));if(result) return result;}}}else{result = false;}return result;}}CrossProcess.java

package com.others;import java.util.ArrayList; import java.util.List;/*** 過河的過程* @author xuefeihu**/ public class CrossProcess {/** 全部步驟 **/private List<CrossStep> steps = new ArrayList<CrossStep>();/*** 加入步驟* @param step 步驟*/public void addStep(CrossStep step){if(step.getDirection()){step.addNextLeftIndex();}else{step.addNextRightIndex();}this.steps.add(step);}/*** 刪除最后一步*/public CrossStep removeLastStep(){return this.steps.remove(this.steps.size()-1);}/*** 獲取最后一個步驟* @return*/public CrossStep getLastStep(){return this.steps.get(this.steps.size()-1);}/*** 打印步驟信息*/public void printStepMessage(){for(CrossStep step : steps){System.out.println(step.getMessage());}}/*** 獲得當前步驟數* @return*/public int getStepCount(){return this.steps.size();}}

CrossStep.java

package com.sunrise.others; /*** 過河步驟* @author xuefeihu**/ public class CrossStep {/** 步驟數 **/private int stepCount;/** 方向:true是左到右，false是右到左 **/private boolean direction;/** 此步驟運送的動物 **/private Animal animal;/** 該步驟之前狀態 **/private Status status;/** 打印語句 **/private String message;/** 下一個左側須要移動的索引 **/private int nextLeftIndex = 0;/** 下一個右側須要移動的索引 **/private int nextRightIndex = 0;/*** 構造器* @param stepCount 步驟數* @param direction 方向:true是左到右，false是右到左* @param animal 此步驟運送的動物* @param status 當前狀態*/public CrossStep(int stepCount, boolean direction, Animal animal, Status status) {this.stepCount = stepCount;this.direction = direction;this.animal = animal;this.status = status;this.message = "第"+stepCount+"步：農夫將" + (this.animal==null?" 自己 ":this.animal.type) + (this.direction ? "從左岸運到右岸" : "從右岸運到左岸");}public int getStepCount() {return stepCount;}public boolean getDirection() {return direction;}public Animal getAnimal() {return animal;}public Status getStatus() {return status;}public String getMessage() {return message;}public int getNextLeftIndex() {return nextLeftIndex;}public void addNextLeftIndex() {this.nextLeftIndex++;}public int getNextRightIndex() {return nextRightIndex;}public void addNextRightIndex() {this.nextRightIndex++;}public void setAnimal(Animal animal) {this.animal = animal;this.message = "第"+stepCount+"步：農夫將" + (this.animal==null?" 自己 ":this.animal.type) + (this.direction ? "從左岸運到右岸" : "從右岸運到左岸");}}
Status.java

package com.sunrise.others;import java.util.ArrayList; import java.util.List;/*** 當前狀態* @author xuefeihu**/ public class Status {/** 左側 **/private List<Animal> left = new ArrayList<Animal>();/** 右側 **/private List<Animal> right = new ArrayList<Animal>();/** 人的位置 **/private boolean flag;/*** 構造狀態對象--克隆對應數據* @param left 左側狀態* @param right 右側狀態* @param flag 人的位置* @param preSelf 前一個動作是否是人單獨過河*/public Status(List<Animal> left, List<Animal> right, boolean flag) {this.left = newList(left);this.right = newList(right);this.flag = flag;}/*** 克隆List對象*/private List<Animal> newList(List<Animal> array){List<Animal> result = new ArrayList<Animal>();for(Animal animal : array){result.add(animal);}return result;}public List<Animal> getLeft() {return left;}public List<Animal> getRight() {return right;}public boolean getFlag() {return flag;}}
Animal.java

package com.sunrise.others; /*** 動物實體* @author xuefeihu**/ public class Animal {public String type = null;public int id = -1;public int afraid = -1;public int control = -1;public Animal(String type, int id, int afraid, int control) {this.type = type;this.id = id;this.afraid = afraid;this.control = control;}/*** 設計一個制約關系：檢查動物能否夠并存* @param animal* @return*/boolean check(Animal animal){return this.afraid == animal.id || this.control == animal.id || animal.afraid == this.id || animal.control == this.id;}@Overridepublic int hashCode() {final int prime = 31;int result = 1;result = prime * result + id;return result;}@Overridepublic boolean equals(Object obj) {if (this == obj)return true;if (obj == null)return false;if (getClass() != obj.getClass())return false;Animal other = (Animal) obj;if (id != other.id)return false;return true;}}
友情提示：（面向對象的思想大致覺得是正確的；因為時間倉促，如有改動代碼，請您附上源代碼與大家共享，謝謝！）

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