人工智能导论八数码算法源程序代码.doc

include
#include
#include

//八数码状态对应的节点结构体
struct Node{
int s[3][3];//保存八数码状态，0代表空格
int f,g;//启发函数中的f和g值
struct Node * next;
struct Node *previous;//保存其父节点
};

int open_N=0; //记录Open列表中节点数目

//八数码初始状态
int inital_s[3][3]={
2,1,3,
8,5,4,
7,0,6
};

//八数码目标状态
int final_s[3][3]={
1,2,3,
8,0,4,
7,6,5
};

//------------------------------------------------------------------------
//添加节点函数入口，方法：通过插入排序向指定表添加
//------------------------------------------------------------------------
void Add_Node( struct Node *head, struct Node *p)
{
struct Node *q;

if(head->next)//考虑链表为空
{ q = head->next;
if(p->f < head->next->f){//考虑插入的节点值比链表的第一个节点值小
p->next = head->next;
head->next = p;
}
else {
while(q->next)//考虑插入节点x，形如a<= x <=b
{
if((q->f < p->f ||q->f == p->f) && (q->next->f > p->f || q->next->f == p->f)){
p->next = q->next;
q->next = p;
break;
}
q = q->next;
}
if(q->next == NULL) //考虑插入的节点值比链表最后一个元素的值更大
q->next = p;
}
}
else head->next = p;
}

//------------------------------------------------------------------------
//删除节点函数入口
//------------------------------------------------------------------------
void del_Node(struct Node * head, struct Node *p )
{
struct Node *q;
q = head;
while(q->next)
{
if(q->next == p){
q->next = p->next;
p->next = NULL;
if(q->next == NULL) return;
// free(p);
}

q = q->next;
}
}

//------------------------------------------------------------------------
//判断两个数组是否相等函数入口
//------------------------------------------------------------------------
int equal(int s1[3][3], int s2[3][3])
{
int i,j,flag=0;
for(i=0; i< 3 ; i++)
for(j=0; j< 3 ;j+)
if(s1[i][j] != s2[i][j]){flag = 1; break;}
if(!flag)
return 1;
else return 0;
}

//------------------------------------------------------------------------
//判断后继节点是否存在于Open或Closed表中函数入口
//------------------------------------------------------------------------
int exit_Nodestruct Node * head,int s[3][3], struct Node *Old_Node)
{
struct Node *q=head->next;
int flag = 0;
while(q)
if(equal(q->s,s)) {
flag=1;
Old_Node->next = q;
return 1;}
else q = q->next;
if(!flag) return 0;
}

//----------------------------------------------------------
//计算p(n)的函数入口
//其中p(n)为放错位的数码与其正确的位置之间距离之和
//具体方法：放错位的数码与其正确的位置对应下标差的绝对值之和
//------------------------------------------------------------------------
int wrong_sum(int s[3][3])
{
int i,j,fi,fj,sum=0;
for(i=0 ; i<3; i++)
for(j=0; j<3; j++)
{
for(fi=0; fi<3; fi++)
for(fj=0; fj<3; fj++)
if((final_s[fi][fj] == s[i][j])){
sum += fabs(i - fi) + fabs(j - fj);
break;
}

}
return sum;
}

//------------------------------------------------------------------------
//获取后继结点函数入口
//检查空格每种移动的合法性，如果合法则移动空格得到后继结点
//-----------------
int get_successor(struct Node * BESTNODE, int direction, struct Node *Successor)//扩展BESTNODE，产生其后继结点SUCCESSOR
{
int i,j,i_0,j_0,temp;
for(i=0; i<3; i++)
for(j=0; j<3; j++)
Successor->s[i][j] = BESTNODE->s[i][j];
//获取空格所在位置
for(i=0; i<3; i++)
for(j=0; j<3; j++)
if(BESTNODE->s[i][j] == 0){i_0 = i; j_0 = j;break;}
switch(direction)
{
case 0: if((i_0-1)>-1 ){
temp = Successor->s[i_0][j_0];
Successor->s[i_0][j_0] = Successor->s[i_0-1][j_0];
Successor->s[i_0-1][j_0] = temp;
return 1;
}
else return 0;

case 1: if((j_0-1)>-1){
temp = Successor->s[i_0][j_0];
Successor->s[i_0][j_0] = Successor->s[i_0][j_0-1];
Successor->s[i_0][j_0-1] = temp;
return 1;
}
else return 0;

case 2: if( (j_0+1)<3){
temp = Successor->s[i_0][j_0];
Successor->s[i_0][j_0] = Successor->s[i_0][j_0+1];
Successor->s[i_0][j_0+1] = temp;
return 1;
}
else return 0;

case 3: if((i_0+1)<3 ){
temp = Successor->s[i_0][j_0];
Successor->s[i_0][j_0] = Successor->s[i_0+1][j_0];
Successor->s[i_0+1][j_0] = temp;
return 1;
}
else return 0;

}

}

//------------------------------------------------------------------------
//从OPen表获取最佳节点函数入口
//------------------------------------------------------------------------
struct Node * get_BESTNODE(struct Node *Open)
{
return Open->next;
}

//------------------------------------------------------------------------
//输出最佳路径函数入口
//------------------------------------------------------------------------
void print_Path(struct Node * head)
{
struct Node *q, *q1,*p;
int i,j,count=1;
p = (struct Node *)malloc(sizeof(struct Node));

//通过头插法变更节点输出次序
p->previous = NULL;
q = head;
while(q)
{
q1 = q->previous;
q->previous = p->previous;
p->previous = q;
q = q1;
}
q = p->previous;

while(q)
{
if(q == p->previous)printf("八数码的初始状态：\n");
else if(q->previous == NULL)printf("八数码的目标状态：\n");
else printf("八数码的中间态%d\n",count++);
for(i=0; i<3; i++)
for(j=0; j<3; j++)
{
printf("%4d",q->s[i][j]);
if(j == 2)printf("\n");
}

printf("f=%d, g=%d\n\n",q->f,q->g);
q = q->previous;
}

}

//-----------------------------------------------
//A*子算法入口:处理后继结点
//------------------------------------------------------------------------
void sub_A_algorithm(struct Node * Open, struct Node * BESTNODE, struct Node * Closed,struct Node *Successor)
{

struct Node * Old_Node = (struct Node *)malloc(sizeof(struct Node));
Successor->previous = BESTNODE;//建立从successor返回BESTNODE的指针
Successor->g = BESTNODE->g + 1;//计算后继结点的g值
//检查后继结点是否已存在于Open和Closed表中，如果存在：该节点记为old_Node，比较后继结点的g值和表中old_Node节点
//g值，前者小代表新的路径比老路径更好，将Old_Node的父节点改为BESTNODE，并修改其f，g值，后者小则什么也不做。
//即不存在Open也不存在Closed表则将其加入OPen表，并计算其f值
if( exit_Node(Open, Successor->s, Old_Node) ){
if(Successor->g < Old_Node->g){
Old_Node->next->previous = BESTNODE;//将Old_Node的父节点改为BESTNODE
Old_Node->next->g = Successor->g;//修改g值
Old_Node->next->f = Old_Node->g + wrong_sum(Old_Node->s);//修改f值
//排序~~~~~~~~~~~~~~~~~~
del_Node(Open Old_Node);
Add_Node(Open, Old_Node);
}

}
else if( exit_Node(Closed, Successor->s, Old_Node)){
if(Successor->g < Old_Node->g){
Old_Node->next->previous = BESTNODE;
Old_Node->next->g = Successor->g;
Old_Node->next->f = Old_Node->g + wrong_sum(Old_Node->s);
//排序~~~~~~~~~~~~~~~~~~
del_Node(Closed, Old_Node);
Add_Node(Closed, Old_Node);
}
}
else {
Successor->f = Successor->g + wrong_sum(Successor->s);
Add_Node(Open, Successor);
open_N++;
}
}

//-----------------------------------------------------------
//A*算法入口
//八数码问题的启发函数为：f(n)=d(n)+p(n)
//其中A*算法中的g(n)根据具体情况设计为d(n)，意为n节点的深度，而h(n)设计为p(n)，
//意为放错的数码与正确的位置距离之和
//------------------------------------------------------------------------
void A_algorithm(struct Node * Open, struct Node * Closed) //A*算法
{
int i,j;
struct Node * BESTNODE, *inital, * Successor;

inital = (struct Node * )malloc(sizeof(struct Node));

//初始化起始节点
for(i=0; i<3; i++)
for(j=0; j<3; j++)
inital->s[i][j] = inital_s[i][j];
inital->f = wrong_sum(inital_s);
inital->g = 0;
inital->previous = NULL;
inital->next = NULL;

Add_Node(Open, inital);//把初始节点放入OPEN表
open_N++;
while(1)
{
if(open_N == 0){printf("failure!"); return;}
else {
BESTNODE = get_BESTNODE(Open);//从OPEN表获取f值最小的BESTNODE，将其从OPEN表删除并加入CLOSED表中
del_Node(Open, BESTNODE);
open_N--;
Add_Node(Closed, BESTNODE);

if(equal(BESTNODE->s, final_s)) {//判断BESTNODE是否为目标节点
printf("success!\n");
print_Path(BESTNODE);
return;
}
//针对八数码问题，后继结点Successor的扩展方法：空格（二维数组中的0）上下左右移动，
//判断每种移动的有效性，有效则转向A*子算法处理后继节点，否则进行下一种移动
else{
Successor = (struct Node * )malloc(sizeof(struct Node)); Successor->next = NULL;
if(get_successor(BESTNODE, 0, Successor))sub_A_algorithm( Open, BESTNODE, Closed, Successor);
Successor = (struct Node * )malloc(sizeof(struct Node)); Successor->next = NULL;
if(get_successor(BESTNODE, 1, Successor))sub_A_algorithm( Open, BESTNODE, Closed, Successor);
Successor = (struct Node * )malloc(sizeof(struct Node)); Successor->next = NULL;
if(get_successor(BESTNODE, 2, Successor))sub_A_algorithm( Open, BESTNODE, Closed, Successor);
Successor = (struct Node * )malloc(sizeof(struct Node)); Successor->next = NULL;
if(get_successor(BESTNODE, 3, Successor))sub_A_algorithm( Open, BESTNODE, Closed, Successor);
}
}
}
}

//------------------------------------------------------------------------
//main()函数入口
//------------------------------------------------------------------------
void main()
{
//------------------------------------------------------------------------
//判断该初始状态是否无解
//------------------------------------------------------------------------
int a[9],b[9];
int x,k,m,n,i;
int t=0;
int r=0;
int *p;

for(x=0;x<8;x++)//将存放初始状态的二维数组转换为一维数组
{
for(m=0;m<3;m++)
{
for(n=0;n<3;n++)
{
p=&inital_s[m][n];
a[x]=*p;
}
}
}

k==0;
for(x=0;x<8;x++)
{
for(i=0;ifinal_s[m][n];
b[x]=*p;
}
}
}

k==0;
for(x=0;x<8;x++)
{
for(i=0;inext = NULL ; Open->previous = NULL;
Closed->next =NULL; Closed->previous = NULL;
A_algorithm(Open, Closed);
}

}

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