Minimum number of steps Nyoj

1#include <cstdio>2#include <cstring>3#include <queue>4 using namespacestd;5 6 intd[9][9];//Mark whether this point accesses and records distances7 intmaze[9][9]={8     1,1,1,1,1,1,1,1,1,9     1,0,0,1,0,0,1,0,1,Ten     1,0,0,1,1,0,0,0,1, One     1,0,1,0,1,1,0,1,1, A     1,0,0,0,0,1,0,0,1, -     1,1,0,1,0,1,0,0,1, -     1,1,0,1,0,1,0,0,1, the     1,1,0,1,0,0,0,0,1, -     1,1,1,1,1,1,1,1,1}; - Const intinf=1<<Ten; -typedef pair<int,int>P; + intN,sx,sy,gx,gy; - intdx[4]={1,0,-1,0},dy[4]={0,1,0,-1};  +  A intBFS () at { -memset (D,0,sizeof(d)); -Queue<p>que; -      for(intI=0;i<9; i++) -          for(intj=0;j<9; j + +) -d[i][j]=INF; in Que.push (P (Sx,sy)); -d[sx][sy]=0; to      +      while(!que.empty ()) -     { theP p=Que.front (); * Que.pop (); $         if(p.first==gx&&p.second==gy)Panax Notoginseng              Break; -          the          for(intI=0;i<4; i++) +         { A             intNx=p.first+dx[i],ny=p.second+Dy[i]; the              +             if(0<=nx&&nx<=8&&0<=ny&&ny<=8&&maze[nx][ny]==0&&d[nx][ny]==inf)//whether the current position can be joined -             { $ Que.push (P (Nx,ny)); $d[nx][ny]=d[p.first][p.second]+1; -             } -         } the     } -     returnD[gx][gy];Wuyi } the  - intMain () Wu { -scanf"%d",&n); About      while(n--) $     { -scanf"%d%d%d%d",&sx,&sy,&gx,&gy); -printf"%d\n", BFS ()); -     } A     return 0; +}

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Challenge Programming Maze Shortest Path template

Minimum number of steps Nyoj