使用O（log（N）（N + M）进行图搜索
[英]Graph search with O(log(N)(N+M)
问题描述
First an explanation: I'm implementing a package manager emulator in java, ie a package manager that just outputs the right things but doesn't actually install anything. 
首先说明一下：我正在用Java实现一个程序包管理器仿真器，即一个程序包管理器，它只输出正确的内容，而实际上并没有安装任何东西。 The thing is I need to search through every package and all it's dependencies to see if they are all installed correctly. 关键是我需要搜索每个软件包及其所有依赖项，以查看它们是否都正确安装。 I've been told that it is possible to get worst-case running time equal to log(N)(N+M), where N is amount of packages and M is the amount of dependencies. 有人告诉我，有可能获得等于log（N）（N + M）的最坏情况下的运行时间，其中N是软件包的数量，M是依赖项的数量。 I believe it is some kind of BFS implementation that needs to be done, maybe just do a BFS to each package and it's dependencies, so if I get to a package that has been checked through some other package's dependencies it will just skip over. 我相信这是需要完成的某种BFS实现，也许只是对每个软件包及其依赖项都进行了BFS调整，因此，如果我找到一个已通过其他软件包的相关性检查过的软件包，它将跳过。 This however, in my head, does not seem to have the right worst-case running time. 但是，在我看来，这似乎没有正确的最坏情况下的运行时间。
Does anyone have any idea on how to achieve getting the right running time? 是否有人对如何获得正确的运行时间有任何想法？
1 个解决方案
解决方案1
0 2015-02-11 18:27:40
I think that this can be done by a topological sort and is often done by piggybacking off of DFS, not BFS. 我认为这可以通过拓扑排序来完成，并且通常可以通过背负DFS（而非BFS）来完成。 That has a linear time of O(M + N). 线性时间为O（M + N）。
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