【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*4],a[MAXN*4];
int n,sum;
void adddata(int now)
{
    segtree[now]=segtree[(now<<1)]+segtree[(now<<1)+1];
}
void buildtree(int now,int l,int r)
{
    segtree[now]=0;
    if (l==r)    return;
    int mid=(l+r)>>1;
    buildtree((now<<1),l,mid);
    buildtree((now<<1)+1,mid+1,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)>>1,ans=0;
    if (begin<=mid)    ans+=query((now<<1),l,mid,begin,end);
    if (end>mid)    ans+=query((now<<1)+1,mid+1,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)>>1;
    if (x<=mid)    pointchange((now<<1),l,mid,x,v);
    else    pointchange((now<<1)+1,mid+1,r,x,v);
    adddata(now);
}
int main()
{
    int i;
    int minn;
    while (~scanf("%d",&n))
    {
        buildtree(1,0,n-1);
        sum=0;
        for (i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            sum+=query(1,0,n-1,a[i],n-1 );
            pointchange(1,0,n-1,a[i],1);
        }
        minn=sum;
        for (i=0;i<n;i++)
        {
            sum+=n-2*a[i]-1;
            minn=min(minn,sum);
        }
        printf("%d
",minn);
    }
    return 0;
}