Minimum Inversion Number

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

Author

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define MAX 5050
using namespace std;
struct Node
{
    int l,r,sum;
}node[MAX<<2];
void build_tree(int l,int r,int root)
{
    node[root].l=l;
    node[root].r=r;
    if(l==r)
    {
        node[root].sum=0;
        return ;
    }
    int mid=(l+r)>>1;
    build_tree(l,mid,root<<1);
    build_tree(mid+1,r,(root<<1)|1);
    node[root].sum=0;
}
void add(int i,int t,int val)
{
    node[i].sum+=val;
    if(node[i].l==node[i].r){
        return ;
    }
    int mid=(node[i].l+node[i].r)>>1;
    if(t<=mid)add(i<<1,t,val);
    else add((i<<1)|1,t,val);
}
int sum(int l,int r,int root)
{
    if(node[root].l==l&&node[root].r==r)
        return node[root].sum;
    int mid=(node[root].r+node[root].l)>>1;
    if(r<=mid)return sum(l,r,root<<1);
    else if(l>mid)return sum(l,r,(root<<1)|1);
    else
    {
        return sum(l,mid,root<<1)+sum(mid+1,r,(root<<1)|1);
    }
}
int a[MAX];
int main()
{
  int n;
  while(~scanf("%d",&n))
  {
      build_tree(1,n,1);
      for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
      int ans=0;
      for(int i=1;i<=n;i++)
      {
          ans+=sum(a[i]+1,n,1);
          add(1,a[i]+1,1);
      }
      int Min=ans;
      for(int i=1;i<=n;i++){
        ans-=a[i];
        ans+=n-a[i]-1;
        if(ans<Min)Min=ans;
      }
      printf("%d
",Min);
  }
  return 0;
}