C++八方向Astar寻路算法大作业

#define col 27
using namespace std;
const int direction[8][2] = { { -1,-1 },{ -1,0 },{ -1,1 },{ 0,-1 },{ 0,1 },{ 1,-
1 },{ 1,0 },{ 1,1 } };// 方向
enum { virable, wall, inOpen, inClose, start, goal };
{
if (tempPoint->parent->r-tempPoint->r == 1 && tempPoint->parent->c == tempPoint->c)
map[tempPoint->parent->r][tempPoint->parent->c] = 4;
if (tempPoint->parent->r-tempPoint->r == -1 && tempPoint->parent->c == tempPoint->c)
1)
map[tempPoint->parent->r][tempPoint->parent->c] = 8;
if (tempPoint->parent->r-tempPoint->r == -1 && tempPoint->parent->c - tempPoint->c ==
1)
map[tempPoint->parent->r][tempPoint->parent->c] = 9;
temp->openPoint = point;
while (openList->next != NULL)
//按从小到大的顺序存储openList;
{
if (point->f < openList->next->openPoint->f)
{
OpenList* tempadd = openList->next;
作业上交时间：最后一节课前发送到课程邮箱。包括源代码，以及运行出来的效果图。
第 1 页，共 9 页
Homework3：
Solution:
1、算法分析： 地图的构建，需要定义一个整型二维数组，障碍位置为 1，可通过点为 0；同样定义坐标 点结构体，以存放每个点的行列 GHF 父结点等信息。 Astar 算法要用到两个表来存放将要搜索的结点和还未搜索的结点，分别是开放列表 openList 和封闭列表 openList，Astar 算法的计算步骤如下： ① S→openList ② S 邻居→openList openList→S→closeList ③ S 邻居有 8 个如何确定先检查哪个？ G 移动成本(10/14) H Manhattan 距离 H=(|e_r-r|+|e_c-c|)*10 F=G+H S=F 最小的点 ④ 重复②③，openList=NULL 未到终点，说明无通路。当 endPoint 在开放表中的 时候，说明已经找到通路，根据每个结点的父结点信息，通过回溯可以找出通 路。 NOTE : 在第②步中检查邻居时，以下几点需要注意： ① 若邻居结点已经在封闭表 closeList 中，可以直接忽略。 ② 若邻居结点不在开放表 openList 和封闭表 closeList 中，计算 GHF 的值，并设置 该结点的父亲结点 ③ 若邻居结点已经在开放表 openList 中，若能得到更小的 G 值则重新计算 GHF 的值 并更新父结点
closeList->closePoint = openList->openPoint; OpenList*temp = openList; openList = openList->next; delete temp; return; } while (closeList->next != NULL) closeList = closeList->next; CloseList*t = new CloseList; t->closePoint = openList->openPoint; t->next = NULL; closeList->next = t;
{
addPointToCloseList(close, open);
if (open == NULL)
{
cout << "未找到出口！地图有误" << endl;
return;
}
}
point*tempPoint = &a[endpoint_r][endpoint_c];
while (tempPoint->parent->flag != start)
map[tempPoint->parent->r][tempPoint->parent->c] = 5;
if (tempPoint->parent->r-tempPoint->r == 0 && tempPoint->parent->c - tempPoint->c == 1)
map[tempPoint->parent->r][tempPoint->parent->c] = 6;
if (tempPoint->parent->r - tempPoint->r == 1&& tempPoint->parent->c - tempPoint->c ==
aStar::aStar() {
第 4 页，共 9 页
steps = 0; startpoint_r = -1; startpoint_c = -1; endpoint_c = -1; endpoint_r = -1; }
void aStar::findGoal(OpenList* open, CloseList* close, int map[row + 2][col + 2])
temp->next = tempadd;
break;
第 3 页，共 9 页
} else
openList = openList->next; } openList->next = temp; } //往封闭表中添加元素 void addPointToCloseList(CloseList*closeList, OpenList*&openList) { if (openList == NULL) { cout << "No data in the OpenList!\n"; return; } if (openList->openPoint->flag != start)openList->openPoint->flag = inClose; if (closeList->closePoint == NULL) {
{
insertToOpenList(open, startpoint_r, startpoint_c);// 起点
addPointToCloseList(close, open);
// 起点放到 close中
while (!insertToOpenList(open, open->openPoint->r, open->openPoint->c))
//定义方格点
struct point
{
int flag; //标志位 0 为可走， 1 为墙壁 2 在penlist 3 在 closelist中 4 为起点 5 为终点
unsigned int r;
unsigned int c;
unsigned int h;
unsigned int g;
unsigned int f;
OpenList*tempOpen = openList; openList = openList->next; delete tempOpen; } //定义aStar类 class aStar { public: point a[row][col]; aStar(); ~aStar() {}
void initPointMap(int map[row + 2][col + 2], OpenList *open); void findGoal(OpenList* open, CloseList* close, int map[row + 2][col + 2]); OpenList* minInOpen(OpenList* open); bool insertToOpenList(OpenList*, int r, int c); bool isInOpenList(OpenList*, int r, int c); bool isInCloseList(OpenList*, int r, int c); void isChangeParent(OpenList*, int r, int c); bool isValid(OpenList*, int r, int c); unsigned int Manhattan(int m_r, int m_c, int r, int c); void counter() { cout << "\n从起点走到终点至少需要" << steps+1 << "步!\n"; cout << endl; } private: unsigned int steps; int startpoint_r; int startpoint_c; int endpoint_r; int endpoint_c; };