商人过河问题C语言源码.c

#include<stdio.h>
#include<stdlib.h>
#include<math.h>             /*double pow(x,y)*/

typedef struct S             /*允许状态集合数据类型*/
{
 int x;               /*商人数*/
 int y;               /*随从数*/
 int boat;              /*船位置标志0此岸,1对岸*/
}S;

typedef struct D             /*允许决策集合数据类型*/
{
 int u;               /*渡船上商人数*/
 int v;               /*渡船上随从数*/
 int flag;              /*5种渡河状态标志*/
                 /*(flag=1:u=1,v=1);(flag=2:u=2,v=0);*/
                 /*(flag=3:u=0,v=2);(flag=4:u=1,v=0);*/
                 /*(flag=5:u=0,v=1)*/
}D;

typedef struct LNode            /*状态链表*/
{
 int lx;
 int ly;
 int lboat;
 int lflag;
 struct LNode *front;
 struct LNode *next;
}LNode,*LinkList;

 

void main()
{
 S s;               /*定义允许状态集合s*/
 LinkList p1,p2,p3,search;
 D d;               /*定义决策状态集合d*/
 int n,k,ok;              /*商人数n,渡河次数k,渡河策略状态allFall*/
 clrscr();
 printf("Displaying System of  Trader & Retinue");
 printf("/n   (C)Copyright  LiliSoftWare 1996-2005");
 printf("/n                       Designed by Lili Wang!/n");
 printf("/nPlease input number of trader:");
 scanf("%d",&s.x);
 s.y=s.x;
 n=s.x;
 s.boat=0;              /*第一次渡河前船在此岸*/
 k=0;  ok=0;  ;          /*初始状态量,allFall=0找到策略allFall=1未找到*/
 p1=(LinkList)malloc(sizeof(LNode));        /*构建状态表*/
 p1->lx=s.x;
 p1->ly=s.y;
 p1->lboat=s.boat;
 p1->lflag=0;
 p1->front=NULL;
 p1->next=NULL;
 p3=p1;

 do
 {
  ok=0;              /*开使新一次渡河*/
  k++;              /*渡河次数加1*/
  d.flag=1;             /*每次都从第一种渡河策略开始判断*/
  do
  {
RELLOW:
   switch(d.flag)
   {
   case 1: d.u=1;d.v=1;break;
   case 2: d.u=2;d.v=0;break;
   case 3: d.u=0;d.v=2;break;
   case 4: d.u=1;d.v=0;break;
   case 5: d.u=0;d.v=1;break;
   default:
     p2=p3->front;free(p3);       /*没有策略符合回到前一次状态*/
     p3=p2;p3->next=NULL;
     if(p3==NULL)
     {
      printf("/nNo answer!");
      getch();
      exit(0);
     }
     else
     {
      s.x=p3->lx;s.y=p3->ly;s.boat=p3->lboat;
      d.flag=p3->lflag;
      d.flag++;
      k--;
      /*goto RELLOW;*/
      continue;
     }

   }/*switch*/
  s.x+=(int)(pow(-1,k))*d.u;
  s.y+=(int)(pow(-1,k))*d.v;
  if(s.boat==0)
   s.boat=1;
  else
   s.boat=0;

  if((s.x==s.y||s.x==0||s.x==n)&&s.x<=n&&s.y<=n&&s.x>=0&&s.y>=0)/*附和允许状态集合*/
  {
   search=p1;

   for(;(search->lx!=s.x||search->ly!=s.y||search->lboat!=s.boat)&&search!=NULL;search=search->next);

   if(search!=NULL)          /*若状态已在表中则为死循环,放弃该策略*/
   {
    d.flag++;
    s.x=p3->lx;           /*恢复s状态*/
    s.y=p3->ly;
    s.boat=p3->lboat;
    continue;
   }
   else
   {
    /*状态不在表中,该策略可行*/
    ok=1;            /*置标志*/
    p2=(LinkList)malloc(sizeof(LNode));     /*将该状态放入允许状态表*/
    p2->next=NULL;
    p2->lx=s.x;
    p2->ly=s.y;
    p2->lboat=s.boat;
    p2->lflag=0;
    p3->next=p2;
    p2->front=p3;
    p3->lflag=d.flag;
    p3=p2;
   }/*else*/
  }/*if*/
  else              /*不在允许状态集合*/
  {
   d.flag++;
   s.x=p3->lx;            /*恢复s状态*/
   s.y=p3->ly;
   s.boat=p3->lboat;

  }

  }while(ok!=1);/*选择每次渡河策略*/

 }while(s.x!=0||s.y!=0);/*全部过河*/

 p3=p1;

 p1=p1->next;
 for(;p1!=NULL;p1=p1->next)
 {
  /*printf("/ntrader:%d,retinue:%d,boat:%d",p1->lx,p1->ly,p1->lboat);*/  /*这是允许状态集合,去掉注释符即可显示*/
  if(p1->lboat==1)
  {
   printf("/nS(trader:%d,retinue:%d):",p1->lx,p1->ly);  /*允许状态集合*/
                 /*渡河策略*/ 
   printf("      trader:%d,retinue:%d   ->",abs((p1->front)->lx-(p1->lx)),abs((p1->front)->ly-(p1->ly)));
  }
  else
  {
   printf("/nS(trader:%d,retinue:%d):",p1->lx,p1->ly);  /*允许状态集合*/
                 /*渡河策略*/
   printf("      trader:%d,retinue:%d   <-",abs((p1->front)->lx-(p1->lx)),abs((p1->front)->ly-(p1->ly)));
  }

  getch();
 }
 printf("/n   Success!!!");
 getch();

 while(p3!=NULL)             /*释放内存*/
 {
  p1=p3;
  p3=p3->next;
  free(p1);
 }

}