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人工智能导论八数码算法源程序代码.doc

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#include<stdio.h> #include<stdlib.h> #include<math.h> //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_Node(struct 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;i<x;i++) { if(a[i]==0) break; else if(a[i]<a[x]) { k++;//k是x所在位置前面比x小的数 } t=t+k; } } for(x=0;x<8;x++)//将存放目标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=&final_s[m][n]; b[x]=*p; } } } k==0; for(x=0;x<8;x++) { for(i=0;i<x;i++) { if(b[i]==0) break; else if(b[i]<b[x]) { k++; } r=r+k; } } if((t+r)%2==1) { printf("该初始状态无解"); return; } else //定义Open和Closed列表。Open列表:保存待检查节点。Closed列表:保存不需要再检查的节点 {struct Node * Open = (struct Node * )malloc(sizeof(struct Node)); struct Node * Closed = (struct Node * )malloc(sizeof(struct Node)); Open->next = NULL ; Open->previous = NULL; Closed->next =NULL; Closed->previous = NULL; A_algorithm(Open, Closed); } } 9
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