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我的打开和关闭列表有问题。如果我改变 B 并移动它，它就不起作用，但是在地图的某些地方，如果我改变 B 的位置，它就会再次起作用。如果有人可以在我出错的地方帮助我，我将不胜感激。谢谢

#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;
}
4

1 回答 1

2

There seem to be a number of things wrong with this:

  1. You never pop from openList, which means that every time you check nodes that you've already looked at, and push nodes into openList that are already in openList or in closedList. This adds a lot of computation.

  2. You have a number of queues and lists when you really only need two. You need one list (or even better, a map so that it's constant time to check whether a node has been already expanded) to keep track of which nodes have been expanded, and you need one priority queue to pull the next node from. So instead of having a nested for loop for the openList, you'd just have the while loop, and then each time you'd pop off the node with the best heuristic value (i.e. the one you expect to get you closest to the goal based on your heuristic).

There are a few more details to work out with the final path, but I think having just two data structures instead of three will push your code in the right direction.

于 2013-04-30T01:30:23.383 回答