Minimum inversion number

Problem descriptionthe 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.
Inputthe 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.
Outputfor each case, output the minimum inversion number on a single line.
Sample Input

101 3 6 9 0 8 5 7 4 2

 
Sample output

16

 
Authorchen, Gaoli
Source

<pre name="code" class="cpp">#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define MAX 5050using 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\n",Min);  }  return 0;}