这篇博客讨论了如何解决一个关于城市交通灯路径的最短时间问题。城市由n×m个交叉路口组成，每个交叉口有一个交通灯，状态为0或1，规定了车辆的移动方向。博主提出使用BFS（广度优先搜索）算法，并通过异或操作处理交通灯状态的变化，来寻找从起点到终点的最短时间。如果无法到达终点，则输出-1。
Description
DreamGrid City is a city with n×mn×m intersections arranged into a grid of nn rows and mm columns. The intersection on the ii-th row and the jj-th column can be described as (i,j)(i,j), and two intersections (i1,j1)(i1,j1) and (i2,j2)(i2,j2) are connected by a road if |i1−i2|+|j1−j2|=1|i1−i2|+|j1−j2|=1.
At each intersection stands a traffic light. A traffic light can only be in one of the two states: 0 and 1. If the traffic light at the intersection (i,j)(i,j) is in state 0, one can only move from (i,j)(i,j) to (i+1,j)(i+1,j) or (i−1,j)(i−1,j); If the traffic light is in state 1, one can only move from (i,j)(i,j)to (i,j+1)(i,j+1) or (i,j−1)(i,j−1) (of course, the destination must be another intersection in the city).
BaoBao lives at the intersection (si,sj)(si,sj), and he wants to visit his best friend DreamGrid living at the intersection (fi,fj)(fi,fj). After his departure, in each minute the following things will happen in order:
- BaoBao moves from his current intersection to another neighboring intersection a
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