(BFS) Poj2935-basic Wall Maze

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The difference between the topic and the most basic BFS maze is that there are some obstacles that can be created by creating a three-dimensional array that marks a place where obstacles cannot go. The other point is the output path, which establishes a pre variable that points to the previous subscript when constructing the struct. Such backtracking (the method is very classical) can be smooth output.

This problem is really very small, but I spent nearly two hours before the smooth AC, is now too low level, to study hard really is a lot of AH. Whatever you do, try your best.

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <queue>5 using namespacestd;6 structGrid7 {8     intx, y;//coordinates9     intPre//to retrace the previous subscriptTen     intdir; One}point[5025]; A intsi,sj,ei,ej,a[4],dir[Ten][Ten][4],direction[4][2]={{-1,0},{0,1},{1,0},{0,-1}},vi[8][8]; - intBFS () - { theMemset (VI,0,sizeof(vi)); -     intFront=0, tail=1; -point[front].x=si; -point[front].y=SJ; +point[front].pre=-1; - grid temp,temp2; +vi[si][sj]=1; A      while(front<tail) at     { -temp=Point[front]; -          for(intI=0;i<4; i++) -         { -             if(dir[temp.x][temp.y][i]==1){Continue;} -             Else in             { -temp2.x=temp.x+direction[i][0]; totemp2.y=temp.y+direction[i][1]; +Temp2.dir=i; -                 if(Vi[temp2.x][temp2.y]) the                     Continue; *                 Else $                 {Panax NotoginsengTemp2.pre=Front; -vi[temp2.x][temp2.y]=1; thepoint[tail++]=Temp2; +                     if(temp2.x==ei&&temp2.y==EJ) A                         return(tail-1); the                 } +             } -         } $front++; $     } - } - voidPrintintX//the classic output mode the { -     if(x>0)Wuyi         { the print (point[x].pre); -         if(point[x].dir==0) Wu         { -printf"N"); About         } $         Else if(point[x].dir==1) -         { -printf"E"); -         } A         Else if(point[x].dir==2) +         { theprintf"S"); -         } $         Elseprintf"W"); the         } the  the } the intMain () - { in      while(SCANF ("%d%d",&sj,&si)) the     { the         intI,j,k,an; About         if(si==0&&sj==0) the              Break; thescanf"%d%d",&ej,&ei); thememset (dir,0,sizeof(dir)); +          for(i=1; i<=6; i++) -         { thedir[1][i][0]=1;Bayidir[6][i][2]=1; thedir[i][1][3]=1; thedir[i][6][1]=1; -         } -          for(j=0;j<3; j + +){ thescanf"%d %d%d%d", &a[1],&a[0],&a[3],&a[2]);//Note that the data is different from the usual the  the         if(a[0]==a[2]) the         { -              for(K=min (a[1],a[3])+1; K<max (a[1],a[3])+1; k++) the             { thedir[a[0]+1][k][0]=1; thedir[a[0]][k][2]=1;94             } the         } the         Else the         {98              for(K=min (a[0],a[2])+1; K<max (a[0],a[2])+1; k++) About             { -dir[k][a[1]+1][3]=1;101dir[k][a[1]][1]=1;102             }103         }104     } thean=BFS ();106 print (an);107Puts"");108     }109     return 0; the}

(BFS) Poj2935-basic Wall Maze