Three-dimensional Sokoban
Problem solving ideas
This is a problem when bfs
analog cuboid various states of
the amount of code a big point of it
AC code
#include<cstdio>
#include<cstring>
using namespace std;
int n,m,x1,y1,px[2500005],py[2500005],pt[2500005],ps[2500005],a[505][505],c[505][505][5];
int dx[4][4]={
{
},{
-1,0,2,0},{
-1,0,1,0},{
-2,0,1,0}};//几种移动
int dy[4][4]={
{
},{
0,2,0,-1},{
0,1,0,-2},{
0,1,0,-1}};
int dt[4][4]={
{
},{
3,2,3,2},{
2,1,2,1},{
1,3,1,3}};
bool check(int x,int y,int t)//判断
{
if(x<1||x>n||y<1||y>m)return false;
if(t==1)
{
if(a[x][y]=='E')return false;
if(a[x][y]=='#')return false;
}
if(t==2)
{
if(a[x][y]=='#')return false;
if(a[x][y-1]=='#')return false;
}
if(t==3)
{
if(a[x][y]=='#')return false;
if(a[x-1][y]=='#')return false;
}
if(c[x][y][t]==1)return false;
return true;
}
void bfs()//bfs
{
int head=0,tail=1;
while(head<tail)
{
head++;
for(int i=0;i<4;i++)
{
int x=px[head]+dx[pt[head]][i],y=py[head]+dy[pt[head]][i],t=dt[pt[head]][i];
if(check(x,y,t))
{
c[x][y][t]=1;
px[++tail]=x;
py[tail]=y;
pt[tail]=t;
ps[tail]=ps[head]+1;
if(x==x1&&y==y1&&t==1)
{
printf("%d\n",ps[tail]);
return;
}
}
}
}
printf("Impossible\n");
return;
}
int main()
{
while(scanf("%d%d",&n,&m))
{
if(n==0&&m==0)return 0;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
a[i][j]=getchar();
while(a[i][j]!='#'&&a[i][j]!='.'&&a[i][j]!='X'&&a[i][j]!='E'&&a[i][j]!='O')
a[i][j]=getchar();
}
memset(c,0,sizeof(c));//预处理，找起点和终点
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
if(a[i][j]=='X')
{
px[1]=i;py[1]=j;ps[1]=0;
if(a[i][j-1]=='X')pt[1]=2;
else if(a[i-1][j]=='X')pt[1]=3;
else pt[1]=1;
c[i][j][pt[1]]=1;
}
if(a[i][j]=='O')
{
x1=i;
y1=j;
}
}
bfs();
}
return 0;
}