hdu1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8178    Accepted Submission(s): 5022

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<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[5005];
struct Node
{
    int l,r,num;
} tree[50000];
void Build(int i,int l,int r)
{
    tree[i].l = l;
    tree[i].r = r;
    tree[i].num = 0;
    if(l == r)
    {
        return;
    }
    int mid = (l+ r) / 2;
    Build(2*i,l,mid);
    Build(2*i+1,mid+1,r);
}

void updata(int i,int x)
{
    int l = tree[i].l;
    int r = tree[i].r;
    int mid = (l + r) / 2;
    if(x == l && x == r)
    {
        tree[i].num ++;
        return;
    }
    if(x <= mid)
        updata(2*i,x);
    else
        updata(2*i+1,x);
    tree[i].num = tree[2*i].num + tree[2*i+1].num;
}

int Query(int n,int x,int y)
{
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    int ans = 0;
    if(x == l && y == r)
        return tree[n].num;
    if(y<= mid)
        ans += Query(2*n,x,y);
    else if(x> mid)
        ans += Query(2*n+1,x,y);
    else
        ans+=Query(2*n,x,mid)+Query(2*n+1,mid+1,y);
    return ans;
}
int main()
{
    int n,sum,ans;
    int i,j;
    while(scanf("%d",&n) != EOF)
    {
        sum = 0;
        Build(1,0,n);
        for(i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            updata(1,a[i]);
            sum += Query(1,a[i]+1,n);
        }
        ans = sum;
        for(i = 1; i <=n; i++)
        {
            sum = sum + (n - 1 - a[i]) - a[i];
            ans=min(ans,sum);
        }
        printf("%d ",ans);
    }
    return 0;
}