HDU 1394 minimum inversion number (line segment tree)

Problem descriptionthe inversion number of a given number sequence A1, A2,..., an is the number of pairs (AI, AJ) that satisfy I <j and AI> AJ.

For a given sequence of numbers A1, A2 ,..., an, if we move the first m> = 0 numbers to the end of the seqence, we will obtain another sequence. there are totally N such sequences as the following:

A1, A2,..., An-1, an (where m = 0-the initial seqence)
A2, A3,..., An, A1 (where M = 1)
A3, A4,..., An, A1, A2 (where m = 2)
...
An, A1, A2,..., An-1 (where M = N-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

Inputthe input consists of a number of test cases. each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the N integers from 0 to n-1.

Outputfor each case, output the minimum inversion number on a single line.

Sample Input

 
101 3 6 9 0 8 5 7 4 2
 

Sample output

 
16
 

Authorchen, Gaoli
Sourcezoj monthly, January 2003

What is the reverse order of the minimum Reverse Order logarithm? This is for two numbers. For example, 2 1 and 2 are in front of 1, while 2 is greater than 1. This is a pair of reverse logarithm. The question is to put the first of the sequence at the end, output with the least logarithm of the reverse order in so many Sequences

First, the data is too big. We should use a line segment tree. The line segment tree stores the number of Le-to-ri values, for example, input 3, we need to query the number of existing numbers from 0 to 3. Let's just look at the code.

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define L (x) (x <1) # define R (X) (x <1 | 1) # define mid (x, y) (x + y)> 1) Using namespace STD; # define n 5005int A [n]; struct stud {int le, Ri; int Va;} f [N * 4]; void build (INT POs, int le, int RI) {f [POS]. le = Le; F [POS]. ri = Ri; F [POS]. va = 0; If (Le = RI) return; int mid = mid (Le, RI); Build (L (POS), Le, mid ); build (R (POS), Mid + 1, RI);} void Update (INT POs, in T le) {f [POS]. va ++; If (F [POS]. le = Le & F [POS]. ri = Le) return; int mid = mid (F [POS]. le, F [POS]. RI); If (mid> = Le) Update (L (POS), Le); else Update (R (POS), Le);} int query (INT POs, int le, int RI) {If (F [POS]. le> = Le & F [POS]. RI <= RI) return f [POS]. va; int mid = mid (F [POS]. le, F [POS]. RI); If (mid> = RI) return query (L (POS), Le, RI); else if (mid <le) return query (R (POS), Le, RI); Return query (L (POS), Le, mid) + query (R (POS), Mid + 1, RI);} int main () {int n, m, I; while (~ Scanf ("% d", & N) {int ans = 0; build (1, 0, n); for (I = 0; I <n; I ++) {scanf ("% d", & A [I]); ans + = query (1, a [I], n-1); Update (1, A [I]);} int temp = ans; for (I = 0; I <n; I ++) {temp = temp-A [I] + N-A [I]-1; if (temp <ans) ans = temp;} printf ("% d \ n ", ans);} return 0 ;}

HDU 1394 minimum inversion number (line segment tree)