修道士与野人渡河问题 数据结构

#include<stdio.h>
void main() 
struct INFO  {  
  int   nSavage; // 岸边野人的数量   开始为3   全部到对岸为0  
  int   nBoanerges; // 岸边传教士的数量   开始为3   全部到对岸为0  
  int   nSide; // 船的位置   在此岸为-1   彼岸为1  
  int   nMoveSavage; // 渡河的野人的数量，用于递归时记录操作状态  
  int   nMoveBoanerges; // 渡河的传教士的数量，用于递归时记录操作状态  
  INFO*   pPrevious;  
  INFO*   pNext;  
  };  
   
  // 判断是否有下一个满足条件的渡河方案  
  bool   GetNextMove(INFO*   pInfo)  
  {  
  // 渡船的优先顺序  
  // 此岸数量多优先，野人优先，彼岸数量少优先，传教士优先。  
  const   int   nStoreCount   =   5;  
  struct   STORE  
  {  
  int   nSavage;  
  int   nBoanerges;  
  };  
   
  STORE   listStore[nStoreCount];  
  if   (pInfo->nSide   ==   -1)  
  {  
  listStore[0].nSavage   =   2;  
  listStore[0].nBoanerges   =   0;  
  listStore[1].nSavage   =   1;  

  listStore[1].nBoanerges   =   1;  
  listStore[2].nSavage   =   0;  
  listStore[2].nBoanerges   =   2;  
  listStore[3].nSavage   =   1;  
  listStore[3].nBoanerges   =   0;  
  listStore[4].nSavage   =   0;  
  listStore[4].nBoanerges   =   1;  
  }  
  else  
  {  
  listStore[0].nSavage   =   0;  
  listStore[0].nBoanerges   =   1;  
  listStore[1].nSavage   =   1;  
  listStore[1].nBoanerges   =   0;  
  listStore[2].nSavage   =   0;  
  listStore[2].nBoanerges   =   2;  
  listStore[3].nSavage   =   1;  
  listStore[3].nBoanerges   =   1;  
  listStore[4].nSavage   =   2;  
  listStore[4].nBoanerges   =   0;  
  }  
   
  int   iStart;  
  if   (pInfo->nMoveSavage   ==   0   &&   pInfo->nMoveBoanerges   ==   0)  
  {  
  iStart   =   0;  
  }  
  else  
  {  
  for   (iStart=0;   iStart<nStoreCount;   iStart++)