Java 动态规划-图论

你在一个大的NxN网格中，有（2这可以通过修改宽度优先的图形搜索来解决。首先，点亮当前房间中所有可能的房间是有意义的。要点亮最大数量的房间，我们必须找到所有可以从

（1,1）

（这意味着有一条路径从（1,1）穿过照亮的房间），然后照亮的房间是所有可以从每个“可到达”房间照亮的房间的联合。我们逐个探索可到达的房间，找到可以照亮的房间。这可能会给我们提供更多可到达的房间

完整的

伪代码是：

queue Q
Q.push (1,1) // room (1,1) reachable
set litupRooms
litupRooms.push(1,1) // room (1,1) is lit up
set visitedRooms
while (Q is not empty)
  room r = Q.pop ()
  if r is in visitedRooms
    continue
  add r to visitedRooms
  for every room that can be lit up from r
    if it is already lit up
      continue
    add it to litupRooms 
    push it to Q if it is adjacent to a room already visited
  for every adjacent room of r
    push it to Q if it is lit up and not visited
return the size of litupRooms

queue Q
Q.push (1,1) // room (1,1) reachable
set litupRooms
litupRooms.push(1,1) // room (1,1) is lit up
set visitedRooms
while (Q is not empty)
  room r = Q.pop ()
  if r is in visitedRooms
    continue
  add r to visitedRooms
  for every room that can be lit up from r
    if it is already lit up
      continue
    add it to litupRooms 
    push it to Q if it is adjacent to a room already visited
  for every adjacent room of r
    push it to Q if it is lit up and not visited
return the size of litupRooms