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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

 

 1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <queue> 6 #include <algorithm> 7 using namespace std; 8  9 struct Point10 {11     char type;12     int x, y, z;13     int steps;14 }start, end;15 16 char map[12][12][12];17 int dir[6][3] = {{1,0,0}, {-1,0,0}, {0,1,0}, {0,-1,0}, {0,0,1}, {0,0,-1}};18 char buff[10];19 int n;20 21 bool islegal(int x, int y, int z)22 {23     return (x>=0 && x<n && y>=0 && y<n && z>=0 && z<n && map[z][x][y]!=X);24 }25 26 void BFS(void)27 {28     queue<Point> qu;29     start.steps = 0;30     qu.push(start);31     while(!qu.empty())32     {33         Point now = qu.front();34         if(now.x==end.x && now.y==end.y && now.z==end.z)35         {36             printf("%d %d\n", n, now.steps);37             return;38         }39         for(int i = 0; i < 6; ++i)40         {41             int xx = now.x + dir[i][0];42             int yy = now.y + dir[i][1];43             int zz = now.z + dir[i][2];44             if(islegal(xx, yy, zz))45             {46                 Point next;47                 next.x = xx, next.y = yy, next.z = zz, next.steps = now.steps + 1;48                 map[zz][xx][yy] = X;49                 qu.push(next);50             }51         }52         qu.pop();53     }54     printf("NO ROUTE\n");55 }56 57 int main(void)58 {59     #ifdef LOCAL60         freopen("1240in.txt", "r", stdin);61     #endif62 63     while(cin >> buff >> n)64     {65         for(int i = 0; i < n; ++i)66             for(int j = 0; j < n; ++j)67                 cin >> map[i][j];68 69         cin >> start.x >> start.y >> start.z;70         cin >> end.x >> end.y >> end.z;71         cin >> buff;72         BFS();73     }74     return 0;75 }
代码君