HDU 1394 Minimum Inversion Number

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

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct segtree
{
    int l,r;
    int num;
    int mid()
    {
        return (l+r)>>1;
    }
};

segtree tree[5005<<2];
int a[5005<<2];

void PushUp(int rt)
{
    tree[rt].num=tree[rt<<1].num+tree[rt<<1|1].num;
}

void build(int l,int r,int rt)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].num=0;
    if(l==r)return;
    int m=tree[rt].mid();
    build(l,m,rt<<1);
    build(m+1,r,rt<<1|1);
}

int query(int l,int r,int rt)
{
    if(tree[rt].l==l&&tree[rt].r==r)
        return tree[rt].num;
    int m=tree[rt].mid();
    if(r<=m)
        return query(l,r,rt<<1);
    else if(l>m)
        return query(l,r,rt<<1|1);
    else
        return query(l,m,rt<<1)+query(m+1,r,rt<<1|1);
}

void update(int pos,int rt)
{
    if(tree[rt].l==tree[rt].r)
    {
        tree[rt].num++;
        return;
    }
    int m=tree[rt].mid();
    if(pos<=m)update(pos,rt<<1);
    else update(pos,rt<<1|1);
    PushUp(rt);
}

int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        build(0,n-1,1);
        int sum=0;
        for(i=0;i<n;i++)
        {
            sum+=query(a[i],n-1,1);
            update(a[i],1);
        }
        int ans=999999999;
        ans=min(ans,sum);
        for(i=0;i<n;i++)
        {
            sum=sum-a[i]+n-1-a[i];
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}