一个人要从一条河的A岸到B岸,他带这,一头狼,一只羊,还有一筐白菜,可是过河的船只能让每次只能让他带这一样东西过河,可是如果,人不在的时候,狼会吃掉羊,羊会吃掉白菜,

可以根据以下步骤编一个程序,让人可以顺利的过河,并且又不损失任何东西,

1,人先带着羊过河,

2,人空手回来,

3,人带着狼过河,

4,人带着羊回来,

5,人带着白菜过河,

6,人空手回来,

7,人带着羊过河

这个问题我前两天发过,可是因为数据库的问题,把我的帖子删了,希望大家能帮帮忙!

// 由于时间的关系，程序没有帮你完全写完，希望有兴趣的朋友把它补充完整

#include <cstdlib> #include <iostream> #include <stdio.h>

using namespace std;

struct Side { char name[50]; int man; int sheep; int wolf; int greens; int situation; int newSituation; void update() { man = situation & 0x8; sheep = situation & 0x4; wolf = situation & 0x2; greens = situation & 0x1; } Side(int s, const char * pn) { strcpy(name, pn); situation = s; update(); } bool check_ok() { int sitation = man | sheep | wolf | greens; bool result = true; for(int i = 5; i<=10; i++) { if(i == sitation) { result = false; break; } } return result; } void report() { // add your code // for example: // *** moved, ***moved } bool move(Side * anotherSide) { bool move; switch(situation) { case 15: situation = 3; anotherSide->situation = 12; update(); anotherSide->update(); report(); anotherSide->report(); move = true; break; case 14:

break;

} return move; } };

int main() { Side SideA(15, "SideA"); Side SideB(0, "SideB");

while(SideA.situation != 0) { if(SideA.move(&SideB)) ; else SideB.move(&SideA); }

system("pause"); return 0; }

程序代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <windows.h>

#define WOLF 3      //狼
#define SHEEP 2     //羊
#define CABBAGE 1   //白菜

int a[3]={WOLF,SHEEP,CABBAGE},b[3];     //A岸，B岸
int count_a=3,count_b=0;                //未过河数，过河数

int main()
{
    int i;      //循环变量
    printf("A岸的初始状态：\t");
    for(i=0;i<count_a;i++)
        printf("%d\t",a[i]);
    printf("\n");
    int state=0;    //往返状态，0为A去往B；1为B返回A
    while(1)
    {
        switch(state)
        {
        case 0:     //A去往B
            if(count_a==3)
                while(1)
                {
                    int t=abs(a[0]-a[1]);
                    if(t==1)
                    {
                        t=a[0];
                        a[0]=a[1];
                        a[1]=a[2];
                        a[2]=t;

                        for(i=0;i<count_a;i++)
                            printf("%d\t",a[i]);
                        printf("\n");
                        continue;
                    }
                    else
                    {
                        b[count_b++]=a[--count_a];
                        printf("\n过河：%d\n",b[count_b-1]);
                        printf("过河后A岸的状态：\t");
                        for(i=0;i<count_a;i++)
                            printf("%d\t",a[i]);
                        printf("\n\n");
                        break;
                    }
                }
            else if(count_a==2)
            {
                int t=a[0];
                a[0]=a[1];
                a[1]=t;
                b[count_b++]=a[--count_a];
                printf("过河：%d\n",b[count_b-1]);
                printf("过河后A岸的状态：\t");
                for(i=0;i<count_a;i++)
                    printf("%d\t",a[i]);
                printf("\n\n");
            }
            else
            {
                b[count_b++]=a[--count_a];
                printf("过河：%d\n",b[count_b-1]);
                printf("过河后A岸的状态：\t");
                for(i=0;i<count_a;i++)
                    printf("%d\t",a[i]);
                printf("\n\n");
            }
            state++;
            system("pause");
            break;
        case 1:     //B返回A
            if(count_b==1)
            {
                printf("返回：人\n返回后的状态：\t");
                for(i=0;i<count_b;i++)
                    printf("%d",b[i]);
                printf("\n\n");
            }
            else if(count_b==2)
            {
                int t=abs(b[0]-b[1]);
                if(t==1)
                {
                    t=b[0];
                    b[0]=b[1];
                    b[1]=t;
                    a[count_a++]=b[--count_b];
                    printf("返回：%d\n",a[count_a-1]);
                }
                else
                    printf("返回：人\n");
                printf("返回后的B岸的状态：\t");
                for(i=0;i<count_b;i++)
                    printf("%d\t",b[i]);
                printf("\n\n");
            }
            state--;
            system("pause");
            break;
        }
        if(count_b==3)
            break;
    }
    printf("成功过河！");
    return 0;
}