Minimum Inversion Number

Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14585 Accepted Submission(s): 8901

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
线段树求逆序数,每当你将开头的a[i],移到尾部的时候,逆序数会减少a[i],也会增加n-a[i]-1,所以变化为n-2*a[i]-1,求最小值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;

const int MAX = 5500;

int Tree[MAX*4];

int a[MAX];

int MM;

int Query(int L,int R,int site,int l,int r)
{
    if(l==L&&R==r)
    {
        return Tree[site];
    }
    int mid=(L+R)>>1;
    if(r<=mid)
    {
        return Query(L,mid,site<<1,l,r);
    }
    else if(l>mid)
    {
        return Query(mid+1,R,site<<1|1,l,r);
    }
    else
    {
        return Query(L,mid,site<<1,l,mid)+Query(mid+1,R,site<<1|1,mid+1,r);
    }
}

void update(int L,int R,int site,int s)
{
    if(L==R)
    {
        Tree[site]++;
        return ;
    }
    Tree[site]++;
    int mid = (L+R)>>1;
    if(mid>=s)
    {
        update(L,mid,site<<1,s);
    }
    else
    {
        update(mid+1,R,site<<1|1,s);
    }
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        memset(Tree,0,sizeof(Tree));
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);//求逆序数
            ans+=Query(0,n-1,1,a[i],n-1);
            update(0,n-1,1,a[i]);
        }
        MM = ans;
        for(int i=1;i<=n;i++)//进行移位
        {
            ans=ans+n-2*a[i]-1;
            if(ans<MM)
            {
                MM=ans;
            }
        }
        printf("%d\n",MM);
    }
    return 0;
}