最佳反转指望INT数组 [英] Optimal inversion counting on int arrays

问题描述

今天，我一直在寻找本地信息学奥林匹克竞赛的最新的考试，我发现了一个有趣的问题。简单地说，它要求为，给定一个整型数组，算多少倒有，在哪里反转是一对indicies的我，Ĵ等>的和 A [1] - ; A [J] 。非正式地，反转的数目是对那些不按顺序的数目。最初，我犯了一个 O（N²）解决方案（是的，天真的），但看到它不会随输入的大小合身，我想过这个问题多一点，然后我意识到，这是可以通过一个变种合并排序，处理好投入的规模在做到这一点为O（n log n）的的时间。

Today I was looking the latest exam of the local informatics olympiad and I found a interesting problem. Briefly, it asks to, given an integer array, count how many inversions it has, where an inversion is a pair of indicies i, j such that i > j and A[i] < A[j]. Informally, the number of inversions is the number of pairs that are out of order. Initially I made a O(n²) solution (yes, the naive one), but seeing it wouldn't fit well with the size of the input, I thought about the problem a bit more and then I realized it's possible to do it within O(n log n) time by a variant of merge sort, which handles good the size of the input.

但在看到输入约束（ N 的整数1和M ，并没有重复），我想知道如果我的解决方案是最佳的，还是你知道还有没有其他办法解决这个问题的跳动为O（n log n）的运行？

But seeing the input constraints (n integers between 1 and M, and no duplicates), I was wondering if my solution is optimal, or do you know if is there any other solution to this problem that beats O(n log n) runtime?

推荐答案

最好的结果文献是为O（n√（log n）的）算法因的Chan和Patrascu 的。不知道有关的常数。

The best result in the literature is an O(n √(log n)) algorithm due to Chan and Patrascu. No idea about the constant.

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