BFS求解迷宮最短路徑

<p>本文使用BFS廣度優先搜索算法實現求解迷宮的最短路徑（C++），使用到了隊列先進先出的性質，依次搜索路徑直到找到目標出口（如果迷宮能走通）求解到的路徑即爲該迷宮的最短路徑，找到返回true，找不到返回false，本文使用vexmap一個map容器記錄隊列的搜索路徑（記錄隊列路徑的實現有點草率，可以再優化）。</p><p> </p><pre class="cpp" name="code">#include<iostream>
#include<queue>
using namespace std;
typedef unsigned long long iUINT64;
map<iUINT64,int> vexmap;
const int g_errNUM = 88888;
offsets myMove4[4] = {//各個方向的偏移表
	{-1,0,"N"},
	{0,1,"E"},
	{1,0,"S"},
	{0,-1,"W"},


};
class  Itmes
{
public:
	iUINT64 x,y;//偏移量x,y和試探方向
	Itmes():x(g_errNUM),y(g_errNUM)
	{
	}
};


bool BFS(Itmes star,Itmes end,int& cntNum)
{
	cntNum = 0;
	iUINT64 curx,cury,searchx,searchy;
	queue<Itmes> ItemQ;
	ItemQ.push(star);
	iUINT64 Star = star.x << 32 | star.y;
	vexMap[Star] = g_errNUM;
	mark[star.x][star.y] = 1;
	while (!ItemQ.empty())
	{
		Itmes tmp = ItemQ.front();
		curx = tmp.x; cury = tmp.y;
		ItemQ.pop();
		for (int i = 0; i < 4; i++)
		{
			searchx = curx + myMove4[i].a;
			searchy = cury + myMove4[i].b;
			if (Maze[searchx][searchy] == 0 && mark[searchx][searchy] == 0)
			{
				mark[searchx][searchy] = 1;
				Itmes tmp2;
				tmp2.x = searchx;
				tmp2.y = searchy;
				iUINT64 NodeCur = curx << 32 | cury;
				iUINT64 NodeSearch = searchx << 32 | searchy;
				vexMap[NodeSearch] = NodeCur;
				ItemQ.push(tmp2);
				//出口條件
				if (searchx == end.x && searchy == end.y)
				{
					iUINT64 Index = searchx << 32 | searchy;
					while (vexMap[Index] != g_errNUM)
					{
						Index = vexMap[Index];
						cntNum++;
						if (cntNum > 1000000)//防護防止出現死循環，可優化
						{
							return false;
						}
					}
					return true;
				}
			}
		}
	}
	return false;
}