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 Description The 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.
Input The 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.
Output For each case, output the minimum inversion number on a single line.
Sample Input

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

Author CHEN, Gaoli Source ZOJ Monthly, January 2003
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1394

題目大意：給一組數，每次可以把開頭的數拿到末尾，問這樣組成的序列中逆序數最小的逆序數值

題目分析：先用樹狀數組求逆序數，每次交換後的逆序數值可以直接算出來，因為是一個0~n-1的排列，所以把第一個數(設為fir)拿到最後先當作減少了fir個逆序數，又這組排列中肯定有n－fir－1個數是大於fir的，再把它們加上，所以逆序數變化的個數為-fir+n-fir-1

#include 
#include 
#include 
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
, mi);
    }
}