BFS求解迷宫最短路径

本文使用BFS广度优先搜索算法实现求解迷宫的最短路径(C++),使用到了队列先进先出的性质,依次搜索路径直到找到目标出口(如果迷宫能走通)求解到的路径即为该迷宫的最短路径,找到返回true,找不到返回false,本文使用vexmap一个map容器记录队列的搜索路径(记录队列路径的实现有点草率,可以再优化)。</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] = {//各个方向的偏移表
 {-1,0,"N"},
 {0,1,"E"},
 {1,0,"S"},
 {0,-1,"W"},


};
class  Itmes
{
public:
 iUINT64 x,y;//偏移量x,y和试探方向
 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);
    //出口条件
    if (searchx == end.x && searchy == end.y)
    {
     iUINT64 Index = searchx << 32 | searchy;
     while (vexMap[Index] != g_errNUM)
     {
      Index = vexMap[Index];
      cntNum++;
      if (cntNum > 1000000)//防护防止出现死循环,可优化
      {
       return false;
      }
     }
     return true;
    }
   }
  }
 }
 return false;
}

 

posted on 2015-12-04 21:23  ykc&boc  阅读(884)  评论(0编辑  收藏  举报