http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8205    Accepted Submission(s): 5041

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 Input101 3 6 9 0 8 5 7 4 2 Sample Output16

#include<stdio.h>
#define maxn 500004
struct node
{
    int left,right;
    int sum;
};
int val[maxn];
int n;
node tree[*maxn];
void build(int left,int right,int i)
{
    tree[i].left =left;
    tree[i].right =right;
    tree[i].sum =;
    if(tree[i].left ==tree[i].right )
        return ;
    build(left,(left+right)/,*i);
    build((left+right)/+,right,*i+);
}
void update(int r,int j)
{
    tree[r].sum++;
    if(tree[r].left==tree[r].right)
        return ;
int mid=(tree[r].left +tree[r].right )/;
      if(j<=mid)
          update(r*,j);
      if(j>mid)
          update(r*+,j);
}
int getsum(int p, int left, int right)
{
    if (left > right)
        return ;
    if (tree[p].left == left && tree[p].right == right)
        return tree[p].sum;
    int m = (tree[p].left + tree[p].right) / ;
    if (right <= m)
        return getsum(p*, left, right);
    else if (left > m)
        return getsum(p*+, left, right);
    else return getsum(p*, left, m) + getsum(p*+, m + , right);
}
int main()
{
    while (~scanf("%d",&n))
    {
        build(,n - ,);
        int i,sum =,ans;
        for (i = ;i<= n; i++)
        {
            scanf("%d", &val[i]);
            sum += getsum(, val[i], n - );
            update(, val[i]);
        }
        ans = sum;
        for (i = ; i <= n; i++)
        {
            sum = sum + (n - val[i] - ) - val[i];
            ans = sum < ans ? sum : ans;
        }
        printf("%d\n", ans);
    }
    return ;
}