HDU 1240 Asteroids! （三维BFS）

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

South Central USA 2001

 

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;
}