标签:输出 post 红色 nts base astar 切换 gets 删除
参考文章:
http://www.policyalmanac.org/games/aStarTutorial.htm 这是英文原文《A*入门》,最经典的讲解,有demo演示
http://www.cnblogs.com/technology/archive/2011/05/26/2058842.html 这是国人翻译后整理的简版,有简单代码demo,不过有些错误,讲得很清晰,本文图片来自这篇
http://blog.csdn.net/b2b160/article/details/4057781 一片关于寻路算法的综述
A*寻路算法是游戏中常用的AI算法,这里用C++简单实现了一下算法,便于理解。
把起始格添加到 "开启列表" do { 寻找开启列表中F值最低的格子, 我们称它为当前格. 把它切换到关闭列表. 对当前格相邻的8格中的每一个 if (它不可通过 || 已经在 "关闭列表" 中) { 什么也不做. } if (它不在开启列表中) { 把它添加进 "开启列表", 把当前格作为这一格的父节点, 计算这一格的 FGH if (它已经在开启列表中) { if (用G值为参考检查新的路径是否更好, 更低的G值意味着更好的路径) { 把这一格的父节点改成当前格, 并且重新计算这一格的 GF 值. } } while( 目标格已经在 "开启列表", 这时候路径被找到) 如果开启列表已经空了, 说明路径不存在. 最后从目标格开始, 沿着每一格的父节点移动直到回到起始格, 这就是路径.
#pragma once /* //A*算法对象类 */ #include <vector> #include <list> const int kCost1=10; //直移一格消耗 const int kCost2=14; //斜移一格消耗 struct Point { int x,y; //点坐标,这里为了方便按照C++的数组来计算,x代表横排,y代表竖列 int F,G,H; //F=G+H Point *parent; //parent的坐标,这里没有用指针,从而简化代码 Point(int _x,int _y):x(_x),y(_y),F(0),G(0),H(0),parent(NULL) //变量初始化 { } }; class Astar { public: void InitAstar(std::vector<std::vector<int>> &_maze); std::list<Point *> GetPath(Point &startPoint,Point &endPoint,bool isIgnoreCorner); private: Point *findPath(Point &startPoint,Point &endPoint,bool isIgnoreCorner); std::vector<Point *> getSurroundPoints(const Point *point,bool isIgnoreCorner) const; bool isCanreach(const Point *point,const Point *target,bool isIgnoreCorner) const; //判断某点是否可以用于下一步判断 Point *isInList(const std::list<Point *> &list,const Point *point) const; //判断开启/关闭列表中是否包含某点 Point *getLeastFpoint(); //从开启列表中返回F值最小的节点 //计算FGH值 int calcG(Point *temp_start,Point *point); int calcH(Point *point,Point *end); int calcF(Point *point); private: std::vector<std::vector<int>> maze; std::list<Point *> openList; //开启列表 std::list<Point *> closeList; //关闭列表 };
Astar.cpp
#include <math.h> #include "Astar.h" void Astar::InitAstar(std::vector<std::vector<int>> &_maze) { maze=_maze; } int Astar::calcG(Point *temp_start,Point *point) { int extraG=(abs(point->x-temp_start->x)+abs(point->y-temp_start->y))==1?kCost1:kCost2; int parentG=point->parent==NULL?0:point->parent->G; //如果是初始节点,则其父节点是空 return parentG+extraG; } int Astar::calcH(Point *point,Point *end) { //用简单的欧几里得距离计算H,这个H的计算是关键,还有很多算法,没深入研究^_^ return sqrt((double)(end->x-point->x)*(double)(end->x-point->x)+(double)(end->y-point->y)*(double)(end->y-point->y))*kCost1; } int Astar::calcF(Point *point) { return point->G+point->H; } Point *Astar::getLeastFpoint() { if(!openList.empty()) { auto resPoint=openList.front(); for(auto &point:openList) if(point->F<resPoint->F) resPoint=point; return resPoint; } return NULL; } Point *Astar::findPath(Point &startPoint,Point &endPoint,bool isIgnoreCorner) { openList.push_back(new Point(startPoint.x,startPoint.y)); //置入起点,拷贝开辟一个节点,内外隔离 while(!openList.empty()) { auto curPoint=getLeastFpoint(); //找到F值最小的点 openList.remove(curPoint); //从开启列表中删除 closeList.push_back(curPoint); //放到关闭列表 //1,找到当前周围八个格中可以通过的格子 auto surroundPoints=getSurroundPoints(curPoint,isIgnoreCorner); for(auto &target:surroundPoints) { //2,对某一个格子,如果它不在开启列表中,加入到开启列表,设置当前格为其父节点,计算F G H if(!isInList(openList,target)) { target->parent=curPoint; target->G=calcG(curPoint,target); target->H=calcH(target,&endPoint); target->F=calcF(target); openList.push_back(target); } //3,对某一个格子,它在开启列表中,计算G值, 如果比原来的大, 就什么都不做, 否则设置它的父节点为当前点,并更新G和F else { int tempG=calcG(curPoint,target); if(tempG<target->G) { target->parent=curPoint; target->G=tempG; target->F=calcF(target); } } Point *resPoint=isInList(openList,&endPoint); if(resPoint) return resPoint; //返回列表里的节点指针,不要用原来传入的endpoint指针,因为发生了深拷贝 } } return NULL; } std::list<Point *> Astar::GetPath(Point &startPoint,Point &endPoint,bool isIgnoreCorner) { Point *result=findPath(startPoint,endPoint,isIgnoreCorner); std::list<Point *> path; //返回路径,如果没找到路径,返回空链表 while(result) { path.push_front(result); result=result->parent; } return path; } Point *Astar::isInList(const std::list<Point *> &list,const Point *point) const { //判断某个节点是否在列表中,这里不能比较指针,因为每次加入列表是新开辟的节点,只能比较坐标 for(auto p:list) if(p->x==point->x&&p->y==point->y) return p; return NULL; } bool Astar::isCanreach(const Point *point,const Point *target,bool isIgnoreCorner) const { if(target->x<0||target->x>maze.size()-1 ||target->y<0&&target->y>maze[0].size()-1 ||maze[target->x][target->y]==1 ||target->x==point->x&&target->y==point->y ||isInList(closeList,target)) //如果点与当前节点重合、超出地图、是障碍物、或者在关闭列表中,返回false return false; else { if(abs(point->x-target->x)+abs(point->y-target->y)==1) //非斜角可以 return true; else { //斜对角要判断是否绊住 if(maze[point->x][target->y]==0&&maze[target->x][point->y]==0) return true; else return isIgnoreCorner; } } } std::vector<Point *> Astar::getSurroundPoints(const Point *point,bool isIgnoreCorner) const { std::vector<Point *> surroundPoints; for(int x=point->x-1;x<=point->x+1;x++) for(int y=point->y-1;y<=point->y+1;y++) if(isCanreach(point,new Point(x,y),isIgnoreCorner)) surroundPoints.push_back(new Point(x,y)); return surroundPoints; }
main.cpp
#include <iostream> #include "Astar.h" using namespace std; int main() { //初始化地图,用二维矩阵代表地图,1表示障碍物,0表示可通 vector<vector<int>> maze={ {1,1,1,1,1,1,1,1,1,1,1,1}, {1,0,0,1,1,0,1,0,0,0,0,1}, {1,0,0,1,1,0,0,0,0,0,0,1}, {1,0,0,0,0,0,1,0,0,1,1,1}, {1,1,1,0,0,0,0,0,1,1,0,1}, {1,1,0,1,0,0,0,0,0,0,0,1}, {1,0,1,0,0,0,0,1,0,0,0,1}, {1,1,1,1,1,1,1,1,1,1,1,1} }; Astar astar; astar.InitAstar(maze); //设置起始和结束点 Point start(1,1); Point end(6,10); //A*算法找寻路径 list<Point *> path=astar.GetPath(start,end,false); //打印 for(auto &p:path) cout<<‘(‘<<p->x<<‘,‘<<p->y<<‘)‘<<endl; system("pause"); return 0; }
标签:输出 post 红色 nts base astar 切换 gets 删除
原文地址:http://www.cnblogs.com/wlzy/p/7096114.html