Asteroids!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4749 Accepted Submission(s): 3048
Problem Description
You‘re in space.
You want to get home.
There are asteroids.
You don‘t want to hit them.
Input
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.
Output
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.
Sample Input
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
Sample Output
1 0 3 4 NO ROUTE
Source
Recommend
zf
原题链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1240
题意：输入n,代表n*n*n的区域，下面有n个n*n的矩阵，代表每一层的情况，接下来输入起点和终点的坐标，如 果存在最短路径，输出n和最短路径的长度，不存在输出“NO ROUTE”；
简单BFS。三维注意坐标的含义！！
AC代码：
#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
char a[15][15][15];
bool vis[15][15][15];
int n;
char ch[10];
int ans;
int sx,sy,sz,ex,ey,ez;
int dir[6][3]= {0,1,0,
1,0,0,
-1,0,0,
0,-1,0,
0,0,1,
0,0,-1
};
struct node
{
int x,y,z,t;
};
bool OK(node no)
{
if(no.x<0||no.x>=n||no.y<0||no.y>=n||no.z<0||no.z>=n)
return false;
return true;
}
int BFS()
{
node now,next;
queue<node>q;
memset(vis,false,sizeof(vis));
vis[sz][sx][sy]=true;
now.x=sx;
now.y=sy;
now.z=sz;
now.t=0;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
if(now.x==ex&&now.y==ey&&now.z==ez)
{
return now.t;
}
//cout<<"********************"<<endl;
for(int i=0; i<6; i++)
{
next.x=now.x+dir[i][0];
next.y=now.y+dir[i][1];
next.z=now.z+dir[i][2];
//cout<<"x="<<next.x<<",y="<<next.y<<",z="<<next.z<<endl;
if(OK(next)&&!vis[next.z][next.x][next.y]&&a[next.z][next.x][next.y]!='X')
{
vis[next.z][next.x][next.y]=true;
next.t=now.t+1;
//cout<<"----------------------------x="<<next.x<<",y="<<next.y<<",z="<<next.z<<endl;
q.push(next);
}
}
}
return -1;
}
int main()
{
while(scanf("%s%d",ch,&n)!=EOF)
{
getchar();
for(int i=0; i<n; i++)
{
for(int j=0; j<n; j++)
{
for(int k=0; k<n; k++)
scanf("%c",&a[i][j][k]);
getchar();
}
}
cin>>sy>>sx>>sz;
cin>>ey>>ex>>ez;
cin>>ch;
ans=BFS();
if(ans!=-1)
cout<<n<<" "<<ans<<endl;
else
cout<<"NO ROUTE"<<endl;
}
return 0;
}