Minimum Inversion Number



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 




#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int mm=22222;
int sum[mm],x[mm];
int i,j,k,n,ans,tmp;
void build(int l,int r,int rt)
{
    sum[rt]=0;
    if(l==r)return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void updata(int p,int l,int r,int rt)
{
    ++sum[rt];
    if(l==r)return;
    int m=(l+r)>>1;
    if(p<=m)updata(p,lson);
    else updata(p,rson);
}
int query(int p,int l,int r,int rt)
{
    if(p<=l)return sum[rt];
    int m=(l+r)>>1,ret=0;
    if(p<=m)ret+=query(p,lson);
    if(m+1<n)ret+=query(p,rson);
    return ret;
}
int main()
{
    while(scanf("%d",&n)!=-1)
    {
        build(0,n-1,1);
        for(tmp=i=0;i<n;++i)
        {
            scanf("%d",&x[i]);
            tmp+=query(x[i],0,n-1,1);
            updata(x[i],0,n-1,1);
        }
        ans=tmp;
        for(i=0;i<n;++i)
        {
            tmp+=n-x[i]-x[i]-1;
            ans=min(ans,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}