BFS Simple Maze

Common Maze:

Enter the end point of the maze and find the shortest path BFS example

Use dist[][] array to record the shortest path of the start point to each point

1#include <iostream>2#include <fstream>3#include <stdio.h>4#include <string.h>5#include <queue>6 using namespacestd;7 8 Const intMaxSize = -;9 Const intINF =0XFFF3;Ten intM,n; One intd[2][4] = { {-1,0,1,0}, {0,1,0, -1} }; A intDist[maxsize][maxsize];//Book Method: Use Dist[m][n] to store the shortest path initial INF indicates unreachable - structPos - { the     intx, y; - }st,ed; - CharMaze[maxsize][maxsize]; -  + BOOLCheckintXinty) - { +     if(X <0|| X >= m | | Y <0|| Y >= N)return false; A     if(Maze[x][y] = ='#'|| Dist[x][y]! = INF)return false;//!=inf said he'd been there before . at     return true; - } - intBFS () - { -      for(inti =0; I < m; i++) -        for(intj =0; J < N; J + +) inDIST[I][J] =INF; -queue<structPos>que; to Que.push (ST); +DIST[ST.X][ST.Y] =0;//myself to myself for 0 -      while(!que.empty ()) the     { *         intNX, NY; $Pos node =Que.front ();Panax Notoginseng Que.pop (); -         //ans + +;//the error here is not that every time the team is in the shortest path, the node debugs can be found . the         if(node.x = = Ed.x && node.y = = ed.y)returnDist[ed.x][ed.y]; +          for(inti =0; I <4; i++) A         { theNX = node.x + d[0][i]; +NY = node.y + d[1][i]; -             if(Check (NX, NY)) $             { $ Pos temp; -temp.x =NX; -TEMP.Y = NY;//It's better to use a pair here . theDist[nx][ny] = Dist[node.x][node.y] +1; - Que.push (temp);Wuyi             } the         } -     } Wu     return-1; - } About  $ intMain () - { -Ifstream Cin ("In.txt"); -Freopen ("In.txt","R", stdin); Ascanf"%d%d", &m, &n); + GetChar (); the      for(inti =0; i < m; i++) -     { $ gets (Maze[i]); the     } the      for(inti =0; I < m; i++) the     { the          for(intj =0; J < N; J + +) -         { in             if(Maze[i][j] = ='S') the             { theSt.x =i; AboutSt.y =J; the             } the             if(Maze[i][j] = ='G') the             { +Ed.x =i; -Ed.y =J; the             }Bayi         } the     } the     intAns =BFS (); -     if(ans = =-1) cout <<"No"<<Endl; -     Elsecout << ans <<Endl; the } the //to fill and memset also need to know is not in the loop fill the //record the shortest path with dist instead of every time out of the team ans++ the //time Complexity State transfer is four directions per lattice up to once O (4*m*n)

BFS Simple Maze