A星算法 [英] A-star algorithm
本文介绍了A星算法的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我在使用我的A-star时遇到问题。它找到从我的点A到B的路径，但不是如果地形更复杂，然后我的Find（）函数似乎不是结束。例如，它在20 x 20阵列上工作，但如果你在底部添加一个正方形（'＃'）到最右边的障碍物/墙壁，那么它会失败。
I'm having issues with my A-star implemention. It does find path from my point A to B but not if the terrain is more 'complex', then my Find() function seems to not be ending. For instance, it does work on the 20 x 20 array here but if you add a square ('#') in the bottom to the most-right obstacle/wall then it fails.
我希望有人能指出我在做的任何错误。这是我的代码：
I hope someone can point out any mistakes I'm doing. Here's my code:
#include <iostream>
#include <string>
#include <cmath>
#include <vector>
#include <utility>
#include <algorithm>
#include <queue>
using namespace std;
class CNode
{
public:
CNode() : xPos(0), yPos(0), travelCost(0) {}
CNode(int x, int y) : xPos(x), yPos(y), travelCost(0) {}
CNode(int x, int y, int cost) : xPos(x), yPos(y), travelCost(cost) {}
inline CNode& operator=(const CNode& target)
{
if (*this != target)
{
xPos = target.xPos;
yPos = target.yPos;
travelCost = target.travelCost;
}
return *this;
}
inline bool operator==(const CNode& target) const
{
return xPos == target.xPos && yPos == target.yPos;
}
inline bool operator!=(const CNode& target) const
{
return !(*this == target);
}
inline bool operator<(const CNode& target) const
{
return target.travelCost < travelCost;
}
int xPos, yPos, travelCost;
};
class CPath
{
public:
typedef vector<CNode> nodeList;
nodeList Find(const CNode& startNode, const CNode& endNode, int mapArray[][20])
{
nodeList finalPath, openList, closedList;
finalPath.push_back(startNode);
openList.push_back(startNode);
closedList.push_back(startNode);
while (!openList.empty())
{
// Check each node in the open list
for (size_t i = 0; i < openList.size(); ++i)
{
if (openList[i].xPos == endNode.xPos && openList[i].yPos == endNode.yPos)
return finalPath;
priority_queue<CNode> nodeQueue;
// Get surrounding nodes
for (int x = -1; x <= 1; ++x)
{
for (int y = -1; y <= 1; ++y)
{
const int current_x = openList[i].xPos + x;
const int current_y = openList[i].yPos + y;
bool alreadyCheckedNode = false;
for (size_t i = 0; i < closedList.size(); ++i)
{
if (current_x == closedList[i].xPos && current_y == closedList[i].yPos)
{
alreadyCheckedNode = true;
break;
}
}
if (alreadyCheckedNode)
continue;
// Ignore current coordinate and don't go out of array scope
if (current_x < 0 || current_x > 20 || current_y < 0 ||current_y > 20 || (openList[i].xPos == current_x && openList[i].yPos == current_y))
continue;
// Ignore walls
if (mapArray[current_x][current_y] == '#')
continue;
const int xNodeDifference = abs(current_x - (openList[i].xPos));
const int yNodeDifference = abs(current_y - (openList[i].yPos));
// Diagonal?
const int direction = xNodeDifference == 1 && yNodeDifference == 1 ? 14 : 10;
const int xDistance = abs(current_x - endNode.xPos);
const int yDistance = abs(current_y - endNode.yPos);
int heuristic = 10 * (xDistance + yDistance);
nodeQueue.push(CNode(current_x, current_y, heuristic));
}
}
if (!nodeQueue.empty())
{
// Add the nearest node
openList.push_back(nodeQueue.top());
finalPath.push_back(nodeQueue.top());
// Put into closed list
while (!nodeQueue.empty())
{
closedList.push_back(nodeQueue.top());
nodeQueue.pop();
}
}
}
}
return finalPath;
}
};
int mapArray[20][20] =
{
{ '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#' },
{ '#', 'A', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '#' },
{ '#', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', 'B', '#' },
{ '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#', '#' },
};
int main(int argc, char** argv)
{
CNode start, end;
for (int width = 0; width < 20; ++width)
{
for (int height = 0; height < 20; ++height)
{
if (mapArray[width][height] == 'A')
{
start.xPos = width;
start.yPos = height;
}
else if (mapArray[width][height] == 'B')
{
end.xPos = width;
end.yPos = height;
}
}
}
CPath pathFinder;
CPath::nodeList n = pathFinder.Find(start, end, mapArray);
for (int i = 0; i < n.size(); ++i)
if (mapArray[n[i].xPos][n[i].yPos] != 'A' && mapArray[n[i].xPos][n[i].yPos] != 'B')
mapArray[n[i].xPos][n[i].yPos] = '*';
for (int height = 0; height < 20; ++height)
{
for (int width = 0; width < 20; ++width)
{
if (width % 20 == 0)
cout << endl;
cout << (char)mapArray[height][width] << " ";
}
}
cin.get();
return 0;
}
推荐答案
节点，只将最上面的一个（最接近目标的那个）放入 openList 中进一步考虑;所有其余的直接进入 closedList ，在那里他们被认为是 alreadyCheckedNode 永远。所以自然地，你的追求者去B直到它卡在一个角落。
When considering the neighbors of a node, you put only the top one (the one closest to the destination) into the openList for further consideration; all the rest go straight into the closedList, where they are considered alreadyCheckedNode forever. So naturally your seeker goes towards B until it gets stuck in a corner.
这篇关于A星算法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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