ACM-三维BFS之Asteroids!——hdu1240

Asteroids!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3161    Accepted Submission(s): 2108

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

 

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1240

这道题就是三维广搜，当时刚做完 胜利大逃亡 这道题，感觉很类似。

于是把这道题代码拿过来改了改就交上去。

发现过不了。

经过YM提醒，发现是i,j,k问题。

输入的N*N*N的图，最慢增长的应该是 z轴的数字。

所以Z轴的应该用i来代替读入，因为三重循环的时候i,j,k中i是增长最慢的。

简单点说：就是把曾经  cin>>map[i][j][k]改为cin>>map[j][k][i]

Why？看我之前说的。

然后普普通通的BFS，0MS  A咯~

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : Asteroids!                  *
*Source: hdu 1240                     *
* Hint  : 三维BFS                     *
***************************************
**************************************/
#include <stdio.h>
#include <string>
#include <string.h>
#include <queue>
#include <iostream>
using namespace std;
int f_x,f_y,f_z,n;
bool ispos;
struct Coor
{
    int x,y,z,step;
};
char mapp[51][51][51];
int dis[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
bool vis[51][51][51];

bool judge(int x,int y,int z)
{
    if(x<0 || y<0 || z<0 || x>=n || y>=n || z>=n)   return 0;
    if(vis[x][y][z]==1 || mapp[x][y][z]=='X')   return 0;
    return 1;
}

int bfs(int x,int y,int z)
{
    queue <Coor> q;
    Coor pre,next;
    int i;

    pre.x=x;
    pre.y=y;
    pre.z=z;
    pre.step=0;
    vis[x][y][z]=1;
    q.push(pre);

    while(!q.empty())
    {
        pre=q.front();
        q.pop();
        if(pre.x==f_x && pre.y==f_y && pre.z==f_z)  {ispos=1;return pre.step;}
        for(i=0;i<6;++i)
        {
            next.x=pre.x+dis[i][0];
            next.y=pre.y+dis[i][1];
            next.z=pre.z+dis[i][2];
            if(judge(next.x,next.y,next.z))
            {
                next.step=pre.step+1;
                vis[next.x][next.y][next.z]=1;
                q.push(next);
            }
        }
    }
    return -1;
}

int main()
{
    string str;
    int i,j,k,rout;
    int s_x,s_y,s_z;
    while(cin>>str>>n)
    {
        for(i=0;i<n;++i)
            for(j=0;j<n;++j)
            {
                // 把回车忽略掉
                scanf("%*");
                for(k=0;k<n;++k)
                    scanf("%c",&mapp[j][k][i]);
            }

        scanf("%d%d%d%d%d%d",&s_x,&s_y,&s_z,&f_x,&f_y,&f_z);
        cin>>str;

        memset(vis,0,sizeof(vis));
        ispos=0;
        rout=bfs(s_x,s_y,s_z);
        if(ispos)   printf("%d %d\n",n,rout);
        else    printf("NO ROUTE\n");
    }
    return 0;
}