农夫过河

本课题要求出完整程序，能够解决下面的问题：

一个农夫带着一只羊，一条狼和一颗白菜想从河的东岸到西岸去。河上仅有一条船。假设他每次只能带一只羊，或者一条狼，或者一颗白菜过河，并且当人不在场时，狼和羊，或羊和白菜不能单独在一起。求出他带一只羊，一条狼和一颗白菜过河的所有办法。狼和羊、羊和白菜不能单独在一起，涉及对象较多，而且运算步骤方法较为复杂，要用程序语言实现，需要将具体实例数字化。针对实现整个过程需要多步，不同步骤中各个事物所处位置不同的情况，没有对先后顺序进行约束，这就需要给各个事物依次进行编号，然后依次试探，若试探成功，进行下一步试探。这就需要使用循环或者递归算法，避免随机盲目运算且保证每种情况均试探到。

#include <stdio.h>struct Condition {                                   定义结构体数组int farmer;int wolf;int sheep;int cabbage;}; struct Condition conditions[100];                          char* action[100];                                      定义一个记录过程的数组 void takeWolfOver(int i)                                  带狼过河{action[i] = "带狼过河";conditions[i + 1].wolf = 1;conditions[i + 1].sheep = conditions[i].sheep;conditions[i + 1].cabbage = conditions[i].cabbage;} void takeWolfBack(int i)                                  带狼回来{action[i] = "带狼回来";conditions[i + 1].wolf = 0;conditions[i + 1].sheep = conditions[i].sheep;conditions[i + 1].cabbage = conditions[i].cabbage;} void takeSheepOver(int i)                                 带羊过河{action[i] = "带羊过河";conditions[i + 1].wolf = conditions[i].wolf;conditions[i + 1].sheep = 1;conditions[i + 1].cabbage = conditions[i].cabbage;} void takeSheepBack(int i)                                 带羊回来{action[i] = "带羊回来";conditions[i + 1].wolf = conditions[i].wolf;conditions[i + 1].sheep = 0;conditions[i + 1].cabbage = conditions[i].cabbage;} void takeCabbageOver(int i)                               带白菜过河{action[i] = "带白菜过河";conditions[i + 1].wolf = conditions[i].wolf;conditions[i + 1].sheep = conditions[i].sheep;conditions[i + 1].cabbage = 1;} void takeCabbageBack(int i)                                带白菜回来{action[i] = "带白菜回来";conditions[i + 1].wolf = conditions[i].wolf;conditions[i + 1].sheep = conditions[i].sheep;conditions[i + 1].cabbage = 0;} void getOverBarely(int i)                                   独自过河{action[i] = "独自过河";conditions[i + 1].wolf = conditions[i].wolf;conditions[i + 1].sheep = conditions[i].sheep;conditions[i + 1].cabbage = conditions[i].cabbage;} void getBackBarely(int i)                                   独自回来{action[i] = "独自回来";conditions[i + 1].wolf = conditions[i].wolf;conditions[i + 1].sheep = conditions[i].sheep;conditions[i + 1].cabbage = conditions[i].cabbage;} void showSolution(int i)                                     显示解决方案{int c;printf("%s\n", "Solution:");for (c = 0; c<i; c++){printf("%d. %s\n", c + 1, action[c]);}printf("%s\n", "Ok! Nice job!");}  void tryOneStep(int i){int c;int j;//check for recursion problem.if (i >= 100)                                  当过程超过100步，报错{printf("%s\n", "Index reached 100. Something is wrong. ");return;}//check for win.if (conditions[i].farmer == 1 &&             当狼，农夫，羊，白菜同时在对岸，完成conditions[i].wolf == 1 &&conditions[i].sheep == 1 &&conditions[i].cabbage == 1){showSolution(i);return;}//check for lose.if ((conditions[i].farmer != conditions[i].wolf && conditions[i].wolf == conditions[i].sheep)||(conditions[i].farmer != conditions[i].sheep && conditions[i].sheep == conditions[i].cabbage))                                 当农夫和狼分开，狼和羊在一起，失败{                                     当农夫和羊分开，羊和白菜在一起，失败return;} for (c = 0; c<i; c++)                            检查重复步骤{if (conditions[c].farmer == conditions[i].farmer&&conditions[c].wolf == conditions[i].wolf&&conditions[c].sheep == conditions[i].sheep&&conditions[c].cabbage == conditions[i].cabbage){return;}}  j = i + 1;if (conditions[i].farmer == 0)                        深度优先{                                                       递归法conditions[j].farmer = 1;getOverBarely(i);tryOneStep(j); if (conditions[i].wolf == 0){takeWolfOver(i);tryOneStep(j);}if (conditions[i].sheep == 0){takeSheepOver(i);tryOneStep(j);}if (conditions[i].cabbage == 0){takeCabbageOver(i);tryOneStep(j);}}else{conditions[j].farmer = 0; getBackBarely(i);tryOneStep(j); if (conditions[i].wolf == 1){takeWolfBack(i);tryOneStep(j);}if (conditions[i].sheep == 1){takeSheepBack(i);tryOneStep(j);}if (conditions[i].cabbage == 1){takeCabbageBack(i);tryOneStep(j);}}} int main(){conditions[0].farmer = 0;conditions[0].wolf = 0;conditions[0].sheep = 0;conditions[0].cabbage = 0;//use recursion to find out how to.tryOneStep(0);}