经典数据结构上机题—答案

nt old;//是否old节点,0非,1是Р};Рstruct linkР{Р node fnode;Р link *next;Р link *pri;Р};Рlink *open,*closed,*bestnode,*essor,*p,*q,*r,*s;Рint maze_flag[7][7]={ {0,1,0,0,0,0,0},Р {0,1,0,1,0,1,0},Р {0,1,0,0,0,1,0},Р {0,1,0,1,0,1,0},Р {0,0,0,1,0,0,0},Р {1,1,0,1,0,1,0},Р {0,0,0,0,0,1,0}};//表示迷宫的数组,0可以走,1不可以走Рnode maze[7][7];Рint judge(node n)//判断函数,判断n节点是否可以走Р{Р if(n.flag==1)Р return(1);Р elseР return(0);Р}Рvoid in_open(node n)//将n节点放入open表Р{Р p=open;Р while(p->next!=open)Р {Р if(n.f>=p->fnode.f)Р {Р p->next->pri=(link *)malloc(sizeof(link));Р p->next->pri->pri=p;Р p=p->next;Р p->pri->next=p;Р p->pri->pri->next=p->pri;Р p=p->pri;Р p->fnode.flag=n.flag;Р p->fnode.f=n.f;Р p->fnode.g=n.g;Р p->fnode.h=n.h;Р p->fnode.x=n.x;Р p->fnode.y=n.y;Р p->fnode.old=n.old;Р p->fnode.sign=n.sign=1;Р }Р elseР p=p->next;