HDOJ 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7428    Accepted Submission(s): 4560

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

 

Recommend

Ignatius.L

 

#include <iostream>

#include <cstdio>

#include <cstring>

#define cls(x) memset((x),0,sizeof((x)));

using namespace std;

int a[5100],x[5100],n,cnt,ans,sum;

int main()

{

while(scanf("%d",&n)!=EOF)

{

    sum=0;cls(a);cls(x);

    for(int i=0;i<n;i++)

    {

        scanf("%d",&a);cnt=0;

        for(int j=0;j<i;j++)

        {

            if(a[j]>a)

                cnt++;

        }

        sum+=cnt;

    }

    for(int i=0;i<n;i++)

    {

        cnt=0;

        for(int j=0;j<n;j++)

        {

            if(a[j]<a&&i!=j)

                cnt++;

        }

        x=cnt;

    }

    ans=sum;

    for(int i=0;i<n;i++)

    {

        sum=sum-2*x+n-1;

        ans=min(sum,ans);

    }

    printf("%d ",ans);

}

    return 0;

}