用段树查找给定数组中最大的一笔子阵定数、组中、最大、用段树
我想找到从给定数组中最大的一笔连续的子阵。我知道发现使用动态规划的使用Kadane算法的概念，最大的一笔连续的子阵方法的O（n）的方法。
但是，这需要大量的时间，如果没有一系列的问题都非常大。有没有办法用段树，因为它是pferred回答它解决了O（日志（n））的时间段查询的最佳选择$ P $来解决它。 谢谢你。
解决方案据我的意见，以贾斯汀的回答，您可以增加一个标准的线段树，实现了 O（日志（N））带O的查询时间（n日志（n））的时间来建立树即插入所有n个元素放到树上。
我们的想法是存储在每个节点 v 不只是一个值，而是三个：
为了执行更新操作的节点 v ，你必须重新计算 MAX_VALUE [V]，left_value [V]，right_value [ V] 。这是非常简单的，我想你自己看着办吧自己 - 有一些情况需要考虑
一个查询操作是在一个基本区段树相似于查询操作。唯一的区别是，在这种情况下，你也必须在 left_value [V] 和 right_value [V] 当计算结果 - 再次，有一些简单的情况下，考虑
我希望你会找出ommited细节。如果没有，让我知道，我会给出更详细的解释。
I wanted to find the largest sum continuous subarray from the given array. I know the O(n) approach of finding the largest sum continuous subarray approach using the concept of dynamic programming using Kadane's algorithm.
But it will take a lot of time if the no of range queries are very large. Is there a way to solve it using Segment-Trees as it is the best option preferred to answer range queries which it solves in O(log(n)) time. Thank you.
解决方案According to my comment to Justin's answer, you can augment a standard segment tree to achieve a O(log(n)) query time with O(n log(n)) time to build the tree i.e. to insert all n elements into the tree.
The idea is to store in every node v not just one value, but three:
In order to perform an update operation for a node v, you have to recompute max_value[v], left_value[v], right_value[v]. This is very straightforward and I think you can figure it out by yourself - there are a few cases to consider.
A query operation is similar to a query operation in a basic segment tree. The only difference is that in this case, you have to consider also the left_value[v] and the right_value[v] while computing a result - again, there are a few easy cases to consider.
I hope that you'll figure out ommited details. If not, let me know and I'll give a more detailed explanation.
