hdu 1240 Asteroids!

1.三维的BFS（语文太差好久才看懂）
2.我发现起点到终点的步数可以直接算出
|start.x-end.x|+|start.y-end.y|+|start.z-end.z|
只要判断能否走到终点
3.每次走过新的点都要记录已经走过
4.记得清空队列
5.要额外判断起点和终点相同的情况，用BFS会 no route。

#include<bits/stdc++.h>
using namespace std;
char room[10][10][10];
#define CHECK(x, y, z) (x<Hx && x>=0 && y<Hy && y>=0 && z<Hz && z>=0)
int dir[6][3]{{-1,0,0},{0,-1,0},{1,0,0},{0,1,0},{0,0,1},{0,0,-1}};
int Hx, Hy, Hz;
int N, num;// num记录步长 
char c;
struct node{
    int x, y, z;
}; 
bool BFS(int dx, int dy ,int dz, int *start, int *end){
    num = 0;
    node star, next;
    queue <node> q;
    while(!q.empty()) q.pop();
    star.x = dx; star.y = dy; star.z = dz;
    q.push(star);
    while(!q.empty()){
        star = q.front();
        q.pop();
        for(int i=0; i<6; ++i){
            next.x = star.x + dir[i][0];
            next.y = star.y + dir[i][1];
            next.z = star.z + dir[i][2];
            if(next.x == end[0] && next.y == end[1] && next.z == end[2]){
                num = abs(next.x - start[0])+abs(next.y - start[1])+abs(next.z - start[2]);
                return 1;
            }
            if(CHECK(next.x, next.y, next.z) && room[next.x][next.y][next.z] == 'O'){
                room[next.x][next.y][next.z] = 'X';
                q.push(next);
            }
        }
    }
    return 0;
}
int main(){
    string s;
    int start[3], end[3];
    while(cin >> s >> N){
        Hx = N; Hy = N; Hz = N;
        for(int Z=0; Z<Hz; ++Z)
            for(int Y=0; Y<Hy; ++Y)
                for(int X=0; X<Hx ;++X)
                    cin >> room[X][Y][Z];
        for(int i=0; i<3; ++i) cin >> start[i];
        for(int i=0; i<3; ++i) cin >> end[i];
        cin >> s;
        bool f = 1;
        for(int i=0; i<3; ++i) 
            if(start[i] != end[i]) f=0;
        if(f) cout << N << ' ' << '0' << endl;
        else if(BFS(start[0], start[1], start[2], start, end))
            cout << N << ' ' << num << endl;
        else cout << "NO ROUTE" << endl;
    }
}