【hdu1394】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

16

 

Author

CHEN, Gaoli

 

Source

ZOJ
Monthly, January 2003

题目大意：求逆序对数目，然后将第一个数字置于队尾在求一遍，直到每个数字都到过队尾的最小逆序对数.

思路：用线段树维护，每次插入统计在其右边的逆序对数和，再用公式推导出每次变换的逆序对数。

attention:注意“0”的存在，被坑了好几次。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #define MAXN 100000
 using namespace std;
 int segtree[MAXN*],a[MAXN*];
 int n,sum;
 void adddata(int now)
 {
     segtree[now]=segtree[(now<<)]+segtree[(now<<)+];
 }
 void buildtree(int now,int l,int r)
 {
     segtree[now]=;
     if (l==r)    return;
     int mid=(l+r)>>;
     buildtree((now<<),l,mid);
     buildtree((now<<)+,mid+,r);
     adddata(now);
 }
 int query(int now,int l,int r,int begin,int end)
 {
     if (begin<=l && end>=r)    return segtree[now];
     int mid=(l+r)>>,ans=;
     if (begin<=mid)    ans+=query((now<<),l,mid,begin,end);
     if (end>mid)    ans+=query((now<<)+,mid+,r,begin,end);
     return ans;
 }
 void pointchange(int now,int l,int r,int x,int v)
 {
     if (l==r)    {segtree[now]+=v;    return;}
     int mid=(l+r)>>;
     if (x<=mid)    pointchange((now<<),l,mid,x,v);
     else    pointchange((now<<)+,mid+,r,x,v);
     adddata(now);
 }
 int main()
 {
     int i;
     int minn;
     while (~scanf("%d",&n))
     {
         buildtree(,,n-);
         sum=;
         for (i=;i<n;i++)
         {
             scanf("%d",&a[i]);
             sum+=query(,,n-,a[i],n- );
             pointchange(,,n-,a[i],);
         }
         minn=sum;
         for (i=;i<n;i++)
         {
             sum+=n-*a[i]-;
             minn=min(minn,sum);
         }
         printf("%d\n",minn);
     }
     return ;
 }