HDU 1240 Asteroids!

三维BFS，要注意行、列、高在三维数组中的位置。

题目链接HDU 1240

题目

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

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 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 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

题目解析

BFS遍历，用数组记录步数，不断更新，找到目的地即可。

AC代码

#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define maxn 1000010
using namespace std;
struct point
{
        int x;
        int y;
        int z;
};
//方向数组
int dx[]={0,0,0,0,1,-1};
int dy[]={1,-1,0,0,0,0};
int dz[]={0,0,1,-1,0,0};
int startx,starty,startz;
int endx,endy,endz;
int n,vis[11][11][11];
char str[11][11][11];
bool cango(int x,int y,int z)
{
        if(x>=0&&x<n&&y>=0&&y<n&&z>=0&&z<n&&!vis[x][y][z])
        {
                return true;
        }
        return false;
}
int bfs(int x,int y,int z)
{
        if(x==endx&&y==endy&&z==endz)
        {
                return 0;
        }
        queue<point> q;
        point p;
        p.x=x,p.y=y,p.z=z;
        q.push(p);
        memset(vis,0,sizeof(vis));
        vis[x][y][z]=1;
        while(!q.empty())
        {
                p=q.front();
                q.pop();
                for(int i=0;i<6;i++)
                {
                        point pp;
                        pp.x=p.x+dx[i],pp.y=p.y+dy[i],pp.z=p.z+dz[i];
                        if(cango(pp.x,pp.y,pp.z)&&str[pp.x][pp.y][pp.z]=='O')
                        {
                                q.push(pp);
                                vis[pp.x][pp.y][pp.z]=vis[p.x][p.y][p.z]+1;
                                if(pp.x==endz&&pp.y==endy&&pp.z==endx)//注意要搞明白这些参数真正的意义
                                {
                                        return vis[pp.x][pp.y][pp.z]-1;
                                }
                        }
                }
        }
        return -1;
}
int main()
{
        string fuck;
        while(cin>>fuck>>n)
        {
                for(int i=0;i<n;i++)
                {
                        for(int j=0;j<n;j++)
                        {
                                for(int k=0;k<n;k++)
                                {
                                        cin>>str[i][j][k];
                                }
                        }
                }
                cin>>startx>>starty>>startz;
                cin>>endx>>endy>>endz;
                cin>>fuck;
                int ans=bfs(startz,starty,startx);
                if(ans==-1)
                {
                        cout<<"NO ROUTE"<<endl;
                }
                else
                {
                        cout<<n<<" "<<ans<<endl;
                }
        }
        return 0;
}