Links: POJ2935
Test instructions
6 X 6 Map lattices and grids may have walls the entire map has three walls for the path from the beginning to the end of the road
The wall in the subject can be understood as the X-direction of a position cannot walk, and a three-dimensional array map[x][y][z] indicates that the z direction of (x, y) cannot go
About the record path you can use a pre array to record the compound value of the previous coordinate of each coordinate in the final reverse direction of the output
Code
#include <iostream> #include <cstdio> #include <cstring> #include <queue>using namespace std; struct node{int x, y,} head;int pre[10][10];int Vis[10][10];bool map[7][7][4]; Realizing the function of the wall int dir[4][2]= {1,0,0,1,-1,0,0,-1};char fx[3][3]; Save Direction name int startx,starty,endx,endy;void BFS () {queue<node>q; Vis[head.x][head.y]=1; Q.push (head); while (!q.empty ()) {Head=q.front (); Q.pop (); if (Head.x==endx&&head.y==endy) {return;} int tx,ty; for (int i=0; i<4; i++) {tx=head.x+dir[i][0]; TY=HEAD.Y+DIR[I][1]; if (tx<1| | ty<1| | tx>6| | ty>6| | vis[tx][ty]| | Map[head.x][head.y][i]) continue; Vis[tx][ty]=1; Pre[tx][ty]=head.x*7+head.y; The load coordinate of the previous coordinate Q.push ({tx,ty}); }}}void output () {int i,j=0; int tx,ty; Char q[49]; fx[1][0]= ' E '; fx[0][1]= ' S '; fx[1][2]= ' W '; fx[2][1]= ' N '; X y plus 1 avoid crossing while (Endx!=startx&&endy!=starty) {TX=PRE[ENDX][ENDY]/7; ty=pre[endx][endy]%7; Decrypt the previous coordinate q[j++]=fx[tx-endx+1][ty-endy+1]; Endx=tx;endy=ty; } for (j=j-2;j>=0;j--) cout<<q[j]; cout<<endl; return;} int main () {while (scanf ("%d%d", &head.y,&head.x) && (head.x!=0)) {scanf ("%d%d", &endy,&e NDX); memset (map,false,sizeof (map)); memset (pre,0,sizeof (pre)); memset (vis,0,sizeof (VIS)); int i,j=3,ax,ay,bx,by,t; while (j--)//Implement wall function {scanf ("%d%d%d%d", &AY,&AX,&BY,&BX); if (ay>by) {t=ay;ay=by;by=t;} if (AX>BX) {t=ax;ax=bx;bx=t;} if (Ay==by) {for (i=ax+1; i<=bx; i++) {map[i][ay][1]=1; Map[i][ay+1][3]=1; } } if (AX==BX) {for (i=ay+1; i<=by; i++) {map[ax][i][0]=1; Map[ax+1][i][2]=1; }}} BFS (); Output (); } return 0;}
POJ2935 Basic Wall Maze BFS Record path