Minimum Inversion Number HDU - 1394（逆序数变形）

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

题意：
给出一个排列，每次可以将开头的一个数移到最后面，最后变成an a1 a2 a3… an-1。求移动过程中的最小逆序数，原序列逆序数也算一种。
思路：
移动过程的逆序数变化其实是确定的。
假设开头是x，那么后面的序列大于x的数y = n - x，小于x的数z = x - 1。
逆序数变化为y - x = n - 2 * x + 1。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>

using namespace std;

const int maxn = 5e3 + 7;
int n;
int a[maxn],c[maxn];

void add(int x,int v)
{
    while(x <= n)
    {
        c[x] += v;
        x += x & (-x);
    }
}

int query(int x)
{
    int res = 0;
    while(x)
    {
        res += c[x];
        x -= x & (-x);
    }
    return res;
}

int main()
{
    while(~scanf("%d",&n))
    {
        memset(c,0,sizeof(c));
        int ans = 0,tmp = 0;
        for(int i =  1;i <= n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            tmp += query(n) - query(a[i]);
            add(a[i],1);
        }
        ans = tmp;
        for(int i = 1;i < n;i++)
        {
            tmp = tmp + n - 2 * a[i] + 1;
            ans = min(ans,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}