1、问题描述
一个农夫在河边要过河，但是他带着一匹狼、一只羊和一颗白菜。他需要用船将这三样东西运至对岸，然而，这艘船的空间有限，只容得下他自己和另一样东西（或狼或羊或白菜）。若他不在场看管的话，狼就会吃羊，羊就会去吃白菜。此人如何才能过河。
2、问题分析
根据题意我们可以得出一下结论：
（1）三样东西必须都过河，但是一次只能载一个；
（2）如果农夫不在场，那么狼会吃羊，羊会吃白菜。
3、算法概述
对于此题，抽象成编程问题来解决。将人、狼、羊、菜在此岸都设为1，当成功渡河后则都为0。用类bankState来表示河岸的状态，用类boatState来表示船上的状态。用time来表示运载的次数。
如此一来，问题转化成判断所有的1是不是都转化为0了，以及其中两个在一起时是否合法的问题了。
4、详细设计
按照人、狼、羊、菜的顺序描述，为了描述方便，假设在河岸A，要过河到达到河岸B。其中，人、狼、羊、菜只能为0或1，其他值都是非法的。
新建类boatState用来描述船上的状态，可知船上的状态只有4种可能，即人自己、人和狼、人和羊、人和菜，用实例化对象boat[4]来表示这四种状态。
新建类bankState用来描述河岸的状态，实例化对象stateA[record]用来记录河岸A的状态，stateB[record]用来记录河岸B的状态，其中record为可能需要的最大次数。
初始化stateA[0]值为{1,1,1,1}，实例化boat[4]，其可取值为{1,0,0,0}、{1,1,0,0}、{1,0,1,0}、{1,0,0,1}用于判断船上状态是否合法。
开始渡河。依次判断是否已经全为0，即是否已经成功渡河。若成功，则打印过河过程中河岸A的所有状态，否则继续判断。判断状态是否合法，状态是否出现过已经，若都通过，则进行渡河操作，开始进行递归。
5、运行结果
6、经验总结
通过这次程序设计，首先，对于类的封装性有了新的认识，因为在编写过程中屡次出现非法访问成员变量，现在终于懂了那么一点了，不再停留于模模糊糊的认识了。其次，对于友元函数也有了一定的认识，因为用到了。对于递归的判断有了新的认识。程序很简单，同时也有很多不足之处，所以发表出来，希望大家能够共同学习，共同进步！
7、源码
#include <iostream>
using namespace std;
class boatState;
class bankState;
class boatState //the state of boat
{
public:
boatState(){}
boatState(int p, int w, int s, int c);
private:
int m_person;
int m_wolf;
int m_sheep;
int m_cabbage;
friend class bankState; //let the bankState access to the boatState
};
boatState::boatState(int p, int w, int s, int c)
{
m_person = p;
m_wolf = w;
m_sheep = s;
m_cabbage = c;
}
class bankState //the state of bank
{
public:
bankState(){}
bankState(int p, int w, int s, int c);
void display(); //output the result to the screen
int compareToEnd(); //confirm whether is the end
int notAllow();
//confirm whether the state is allowed
int compareToAppear(bankState temp);
//confirm the state has never appeared
void newState(bankState &tempState, boatState tempBoat);
//the new state of bankA
void otherState(bankState &tempB); //the state of bankB
private:
int m_person;
int m_wolf;
int m_sheep;
int m_cabbage;
};
bankState::bankState(int p, int w, int s, int c)
{
m_person = p;
m_wolf = w;
m_sheep = s;
m_cabbage = c;
}
void bankState::display()
{
cout << m_person << m_wolf << m_sheep << m_cabbage << endl;
}
int bankState::compareToEnd()
{
if(m_person == 0 && m_wolf == 0 && m_sheep == 0 && m_cabbage == 0)
{
return 1;
}
else
{
return 0;
}
}
int bankState::notAllow()
{
if (m_person == 0 && ((m_wolf == 1 && m_sheep == 1 && m_cabbage == 1 )
|| (m_wolf == 1 && m_sheep == 1) || (m_sheep == 1 && m_cabbage == 1)))
{
return 1;
}
else
{
if ((m_person != 0 && m_person != 1) || (m_wolf != 0 && m_wolf != 1)
|| (m_sheep != 0 && m_sheep != 1) || (m_cabbage != 0 && m_cabbage != 1))
{
return 1;
}
else
{
return 0;
}
}
}
int bankState::compareToAppear(bankState temp)
{
if (m_person == temp.m_person && m_wolf == temp.m_wolf
&& m_sheep == temp.m_sheep && m_cabbage == temp.m_cabbage)
{
return 1;
}
else
{
return 0;
}
}
void bankState::newState(bankState &tempState, boatState tempBoat)
{
if (m_person == 1)
{
tempState.m_person = m_person - tempBoat.m_person;
tempState.m_wolf = m_wolf - tempBoat.m_wolf;
tempState.m_sheep = m_sheep - tempBoat.m_sheep;
tempState.m_cabbage = m_cabbage - tempBoat.m_cabbage;
}
if (m_person == 0)
{
tempState.m_person = m_person + tempBoat.m_person;
tempState.m_wolf = m_wolf + tempBoat.m_wolf;
tempState.m_sheep = m_sheep + tempBoat.m_sheep;
tempState.m_cabbage = m_cabbage + tempBoat.m_cabbage;
}
}
void bankState::otherState(bankState &tempB)
{
tempB.m_person = 1 - m_person;
tempB.m_wolf = 1 - m_wolf;
tempB.m_sheep = 1 - m_sheep;
tempB.m_cabbage = 1 - m_cabbage;
}
const int record = 20;
bankState stateA[record];
bankState stateB[record];
boatState boat[4];
void crossRiver(int time)
{
if (stateA[time].compareToEnd())
{
cout << "Success!" << endl;
for (int num = 0; num <= time; num++)
{
stateA[num].display();
}
return;
}
if (stateA[time].notAllow() || stateB[time].notAllow())
{
return;
}
for (int i = 0; i < time; i++)
{
if (stateA[time].compareToAppear(stateA[i]))
{
return;
}
}
for (int j = 0; j < 4; j++)
{
stateA[time].newState(stateA[time + 1], boat[j]);
stateA[time + 1].otherState(stateB[time + 1]);
crossRiver(time + 1);
}
}
void main()
{
//init the state
stateA[0] = bankState(1, 1, 1, 1);
boat[0] = boatState(1, 0, 0, 0);
boat[1] = boatState(1, 1, 0, 0);
boat[2] = boatState(1, 0, 1, 0);
boat[3] = boatState(1, 0, 0, 1);
crossRiver(0);
system("pause");
} |
