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 longusing 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;}