Question

link

2d array ＊代表障碍物 ＃代表货物 空白就是正常的路

问如何找到一个点为出发点 能实现总共取货路径最短？ 每次只能拿一个货物，遇到障碍需要绕开，拿到以后要放回出发点，然后再取另一个.

＊＊＊＊＊＊＊＊＊＊
＊  #           ＊
＊  ＊＊＊  ＊   ＊
＊              ＊
＊     ＊＊   ＊ ＊
＊  #    # # ＊＊＊
＊＊＊＊＊＊＊＊＊＊

Solution

This looks like a very difficult question, especially during a phone interview.

The 10th floor gives the best solution:

BFS from every box. in each box, a non-blocking cell (include box position, but exclude hazard position) will have a weight value, stand for the distance to the box.

after bfs from all the boxes, each cell will have k weight, k is the number of boxes. sum all the weight in each cell, and find the cell with smallest sum of weight.

One problem of this solution may lead to a cell of a box. We can then sort the cell by sum of weight and find the first position that is not a box.

complexity O(k*n^2)

Suggested at this post, we can also use A* search (with heuristic function) to solve.

Code

not written