LeetCode 1192. Critical Connections in a Network 中文解释 Chinese Version

[@hz4353]

难题的救星

[@hongchen9833]

I have seen several videos and docs for this issue. I think you did a great job to clearly explained the problem. Thanks a lot. Also subscribed!

[@happygirlzt]

Glad it was helpful! :D Thank you for subscribing. ;)

[@yilinxie8086]

Parents can just be an int to pass into the recursion? I guess that would save some memory.

[@yvaineliu800]

中文版讲解比较清楚！支持！ 想问一下，从7:44开始解释的那个else if的情况，为什么low[u] = min(low[u], disc[v]), 而不是low[u] = min(low[u], low[v])？

[@happygirlzt]

我的理解啊，我们要尽可能使low[u]更小；如果v已经访问过了,那么之前是比较过low[v]的值的（就是在if (disc[v] == -1) 那个分支）。而由于之前访问过v，所以disc[v]有可能比low[u]小，所以与之比较来更新当前的low[u]，更新了low[u]的值

[@xiuwenzhong7375]

@@happygirlzt 我觉得这里应该是说反了， low[v] 的最大值只可能是 dfn[v], 然后随着对v点处理，low[v]可能会小于dfn[v],但是绝对不会超过dfn[v]。 所以，判断 if (parent[u] != v) 也就是当判断v已经访问过的情况下，我们要更新low[u]，应该使用 min(low[u], low[v]) 这样才能保证 low[u]被update到一个更小的值。 case: 10 [[1,0],[2,0],[3,0],[4,1],[5,3],[6,1],[7,2],[8,1],[9,6],[9,3],[3,2],[4,2],[7,4],[6,2],[8,3],[4,0],[8,6],[6,5],[6,3],[7,5],[8,0],[8,5],[5,4],[2,1],[9,5],[9,7],[9,4],[4,3]]

[@dalindu7572]

这题很难 不好讲！ 加油！

[@rliu6821]

小姐姐，我有点听不太懂。

[@kyriewang9130]

支持！

[@avseqzxcv]

HI, I love your video. May you also analyze time and space complexity in your video in the future?

[@happygirlzt]

! I will try! I am lazy sometimes..

[@yongli1226]

墙裂支持。