HDU 1240 (简单三维广搜) Asteroids!

给出一个三维的迷宫以及起点和终点，求能否到大终点，若果能输出最短步数

三维的问题无非就是变成了6个搜索方向

最后强调一下xyz的顺序，从输入数据来看，读入的顺序是map[z][x][y]

总之，这是很基础的一道题

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <algorithm>
 using namespace std;

 struct Point
 {
     char type;
     int x, y, z;
     int steps;
 }start, end;

 char map[][][];
 int dir[][] = {{,,}, {-,,}, {,,}, {,-,}, {,,}, {,,-}};
 char buff[];
 int n;

 bool islegal(int x, int y, int z)
 {
     return (x>= && x<n && y>= && y<n && z>= && z<n && map[z][x][y]!='X');
 }

 void BFS(void)
 {
     queue<Point> qu;
     start.steps = ;
     qu.push(start);
     while(!qu.empty())
     {
         Point now = qu.front();
         if(now.x==end.x && now.y==end.y && now.z==end.z)
         {
             printf("%d %d\n", n, now.steps);
             return;
         }
         for(int i = ; i < ; ++i)
         {
             int xx = now.x + dir[i][];
             int yy = now.y + dir[i][];
             int zz = now.z + dir[i][];
             if(islegal(xx, yy, zz))
             {
                 Point next;
                 next.x = xx, next.y = yy, next.z = zz, next.steps = now.steps + ;
                 map[zz][xx][yy] = 'X';
                 qu.push(next);
             }
         }
         qu.pop();
     }
     printf("NO ROUTE\n");
 }

 int main(void)
 {
     #ifdef LOCAL
         freopen("1240in.txt", "r", stdin);
     #endif

     while(cin >> buff >> n)
     {
         for(int i = ; i < n; ++i)
             for(int j = ; j < n; ++j)
                 cin >> map[i][j];

         cin >> start.x >> start.y >> start.z;
         cin >> end.x >> end.y >> end.z;
         cin >> buff;
         BFS();
     }
     return ;
 }

代码君