You‘re in space.
You want to get home.
There are asteroids.
You don‘t want to hit them.
Input
to this problem will consist of a (non-empty) series of up to 100 data
sets. Each data set will be formatted according to the following
description, and there will be no blank lines separating data sets.
A single data set has 5 components:
Start line - A single line, "START N", where 1 <= N <= 10.
Slice
list - A series of N slices. Each slice is an N x N matrix representing
a horizontal slice through the asteroid field. Each position in the
matrix will be one of two values:
‘O‘ - (the letter "oh") Empty space
‘X‘ - (upper-case) Asteroid present
Starting
Position - A single line, "A B C", denoting the <A,B,C>
coordinates of your craft‘s starting position. The coordinate values
will be integers separated by individual spaces.
Target Position -
A single line, "D E F", denoting the <D,E,F> coordinates of your
target‘s position. The coordinate values will be integers separated by
individual spaces.
End line - A single line, "END"
The
origin of the coordinate system is <0,0,0>. Therefore, each
component of each coordinate vector will be an integer between 0 and
N-1, inclusive.
The first coordinate in a set indicates the column. Left column = 0.
The second coordinate in a set indicates the row. Top row = 0.
The third coordinate in a set indicates the slice. First slice = 0.
Both the Starting Position and the Target Position will be in empty space.
For each data set, there will be exactly one output set, and there will be no blank lines separating output sets.
A
single output set consists of a single line. If a route exists, the
line will be in the format "X Y", where X is the same as N from the
corresponding input data set and Y is the least number of moves
necessary to get your ship from the starting position to the target
position. If there is no route from the starting position to the target
position, the line will be "NO ROUTE" instead.
A move can only be
in one of the six basic directions: up, down, left, right, forward,
back. Phrased more precisely, a move will either increment or decrement a
single component of your current position vector by 1.
START 1
O
0 0 0
0 0 0
END
START 3
XXX
XXX
XXX
OOO
OOO
OOO
XXX
XXX
XXX
0 0 1
2 2 1
END
START 5
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
XXXXX
XXXXX
XXXXX
XXXXX
XXXXX
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
0 0 0
4 4 4
END
#include<bits/stdc++.h>
using namespace std;
int f[3][6]={1,-1,0,0,0,0,0,0,1,-1,0,0,0,0,0,0,1,-1},vis[20][20][20],mp[20][20][20],n,i,j,k,sx,sy,sz,px,py,pz,anss;
char x;
string s;
int main()
{
while(cin>>s>>n)
{
anss=-1;
memset(vis,0,sizeof(vis));
queue<int>q;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
for(k=0;k<n;k++)
{
cin>>x;
if(x==‘O‘)
mp[j][k][i]=1;
else
mp[j][k][i]=0;
}
}
}
cin>>sx>>sy>>sz;
cin>>px>>py>>pz;
q.push(sx);
q.push(sy);
q.push(sz);
q.push(0);
vis[sx][sy][sz]=1;
while(!q.empty())
{
int x=q.front();
q.pop();
int y=q.front();
q.pop();
int z=q.front();
q.pop();
int step=q.front();
q.pop();
if(x==px&&y==py&&z==pz)
{
anss=step;
break;
}
for(i=0;i<6;i++)
{
int xx=x+f[0][i];
int yy=y+f[1][i];
int zz=z+f[2][i];
if(xx>=0&&xx<n&&yy>=0&&yy<n&&zz>=0&&zz<n&&mp[xx][yy][zz]==1&&vis[xx][yy][zz]==0)
{
q.push(xx);
q.push(yy);
q.push(zz);
q.push(step+1);
vis[xx][yy][zz]=1;
}
}
}
cin>>s;
if(anss!=-1)
{
cout<<n<<‘ ‘<<anss<<endl;
}
else
{
cout<<"NO ROUTE"<<endl;
}
}
}
