人工智能圆盘问题程序.doc

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1、设有3个大小不等的圆盘A,B,C套在一根轴上，每个圆盘上都标有数字1、2、3、4，并且每个圆盘都可以独立的绕轴做逆时针转动，每次转动90度，初始状态S0和目标状态Sg，用深度优先搜索和广度优先搜索求从S0到Sg的路径。代码和结果：代码如下：#include#include"stdio.h"typedefstructTNode{__int64data;TNode*LChild;TNode*RChild;}node;//把所以状态列出来，以右边为基准，则初态为，终态为__int64Node[64]={1234

2、12341234,123412344123,123412343412,123412342341,123441231234,123441234123,123441233412,123441232341,123434121234,123434124123,123434123412,123434122341,123423411234,123423414123,123423413412,123423412341,341212341234,341212344123,341212343412,341212342341,341241231

3、234,341241234123,341241233412,341241232341,341234121234,341234124123,341234123412,341234122341,341223411234,341223414123,341223413412,341223412341,412312341234,412312344123,412312343412,412312342341,412341231234,412341234123,412341233412,412341232341,412334121234,4

4、12334124123,412334123412,412334122341,412323411234,412323414123,412323413412,412323412341,234112341234,234112344123,234112343412,234112342341,234141231234,234141234123,234141233412,234141232341,234134121234,234134124123,234134123412,234134122341,234123411234,234123

5、414123,234123413412,234123412341};//在空二叉树或二叉树中插入一个新的结点node*Insert(node*root,__int64value){node*current;node*parent;node*new_node;new_node=(node*)malloc(sizeof(node));new_node->data=value;new_node->LChild=NULL;new_node->RChild=NULL;if(root==NULL){returnnew_node;}els

6、e{current=root;while(current!=NULL){parent=current;if(current->data>value)current=current->LChild;elsecurrent=current->RChild;}if(parent->data>value)parent->LChild=new_node;elseparent->RChild=new_node;}returnroot;}//创建一棵新二叉树node*Create(__int64*data,intlen){node*roo

7、t=NULL;inti;for(i=0;idata!=412312342341){printf("%I64u--->",root->data);PreOrder(root->LChild);PreOrder(root->RChild);}}//广度优先搜索voidLevel(node*root){constintmaxsize=

8、64;node*Queen[maxsize];node*current=NULL;intrear=0;intfore=0;if(root!=NULL){rear=(rear+1)%maxsize;Queen[rear]=root;}while(fore!=rear){fore=(fore+