HDU 1394 Minimum Inversion number (tree-like array for reverse order)

Minimum Inversion number Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others) Total Submission (s): 13942 Accepted Submission (s): 8514
problem DescriptionThe inversion number of a given number sequence A1, A2, ..., 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'll Obtain another sequence. There is 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 is 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 the 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 in a single line.
 Sample Input

101 3 6 9 0 8 5 7 4 2

 Sample Output

16

 AuthorCHEN, Gaoli SourceZOJ Monthly, January 2003
Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1394

Title: To a group of numbers, each time you can get the beginning of the number to the end, ask such a sequence of the number of reverse order of the smallest inverse value

Title Analysis: First Use the tree-like array to reverse the number, the reverse value after each Exchange can be directly calculated, because it is a 0~n-1 arrangement, so the first number (set to FIR) to the last to reduce the number of FIR in reverse order, and this group is sure to have n-fir-1 number is greater than fir, Add them again, so the number of reverse order numbers is-fir+n-fir-1

#include <cstdio> #include <cstring> #include <algorithm>using namespace std;int const MAX = 5005;int c[ MAX], A[max], n;int lowbit (int x) {    return x & (-X);} void Add (int x) {for    (int i = x; i <= n; i + = Lowbit (i))        C[i] + +;} int Sum (int x) {    int res = 0;    for (int i = x; i > 0; I-= Lowbit (i))        res + = C[i];    return res;} int main () {while    (scanf ("%d", &n)! = EOF)    {        memset (c, 0, sizeof (c));        int ini = 0;        for (int i = 1; I <= n; i++)        {            scanf ("%d", &a[i]);            A[i] + +;            INI + = SUM (n)-sum (A[i]);            ADD (A[i]);        }        int mi = ini;        for (int i = 1; I <= n; i++)        {            ini + = (-(A[i]-1) + N-(A[i]-1)-1);            mi = min (mi, ini);        }        printf ("%d\n", MI);    }}

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HDU 1394 Minimum Inversion number (tree-like array for reverse order)