HDOJ  1240   Asteroids!

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

题目解析：题目给了一个三维的空间坐标系，找一个点到另一个的最短的距离，这其实和二维的一样，

只不过有些的小小的改动，我自己是按照广搜的，每一次搜完之后清空

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
const int MAX=1000000;
using namespace std;
typedef struct point
{
    intx,y,z;
    intstep;
};
char space[13][13][13];
intdri[6][3]={{0,0,1},{0,0,-1},{1,0,0},{-1,0,0},{0,1,0},{0,-1,0}};
queue<point> q;
int A,B,C,D,E,F;
int N;
int summin[13][13][13];
int BFS()
{
    intaa,bb,cc,i;
    pointhd;
   hd.x=A,hd.y=B,hd.z=C;
   hd.step=0;
   q.push(hd);
   while(!q.empty())
    {
       point tmp;
       tmp=q.front();
       //cout<<tmp.x<<tmp.y<<tmp.z<<endl;
       q.pop();
       for(i=0;i<6;i++)
       {
           aa=tmp.x+dri[i][0],bb=tmp.y+dri[i][1],cc=tmp.z+dri[i][2];
           if(aa>=0&&aa<N&&bb>=0&&bb<N&&cc>=0&&cc<N&&space[aa][bb][cc]=='O')
           {
               point t;
               t.x=aa;
               t.y=bb;
               t.z=cc;
               t.step=tmp.step+1;
               if(t.step<summin[aa][bb][cc])
               {
                   q.push(t);
                   summin[aa][bb][cc]=t.step;
               }
           }
       }
    }
    returnsummin[D][E][F];
}
int main()
{
    chara[6],b[4];
    inti,j,k,temp;
   while(scanf("%s%d",a,&N)!=EOF)
    {
       for(i=0;i<N;i++)
           for(j=0;j<N;j++)
               for(k=0;k<N;k++)
               {
                   scanf(" %c",&space[k][j][i]);
                   summin[k][j][i]=MAX;
               }
       cin>>A>>B>>C>>D>>E>>F;
       scanf("%s",b);
       if(A==D&&B==E&&C==F)
       {
           printf("%d %d\n",N,0);
           continue;
       }
       temp=BFS();
       if(temp==MAX)
           printf("%s\n","NO ROUTE");
       else
           printf("%d %d\n",N,temp);
       while(!q.empty())
           q.pop();
    }
    return0;
}