Reduce inversion count

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array. If such a pair exists, return the new inversion count; otherwise returns the original inversion count.

Definition of Inversion: Let (A[0], A[1] ... A[n], n <= 50) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A.

Example:
Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2
InversionCountOfSwap({3, 1, 2})=>
{
 InversionCount({1, 3, 2}) = 1 <-- swapping 1 with 3, decreases inversion count by 1
 InversionCount({2, 1, 3}) = 1 <-- swapping 2 with 3, decreases inversion count by 1
 InversionCount({3, 2, 1}) = 3 <-- swapping 1 with 2 , increases inversion count by 1
}

Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.

Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.

样例输入

3,1,21,2,3,4,5

样例输出

10

我的java代码。。。求大神不吝赐教。

package com.summer.intern;
import java.util.Scanner;
public class Main {
    public static int getInversionCount(int data[])
    {
        int count=0;
        for(int i=0;i<data.length;i++){
            for(int j=i+1;j<data.length;j++)
            {
                if(data[i]>data[j]) count++;
            }
        }
        return count;
    }
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        while (input.hasNextLine()) {
            String line = input.nextLine();
            String num[] = line.split(",");
            int data[]=null;
            data=new int [num.length];
            for(int i=0;i<num.length;i++)
            {
                data[i]=Integer.parseInt(num[i]);
            }
            int orial_inversion_count=getInversionCount(data);
            int temp_data[]=data.clone();
            for(int i=0;i<data.length;i++){
                for(int j=i+1;j<data.length;j++){
                    int temp=temp_data[i];
                    temp_data[i]=temp_data[j];
                    temp_data[j]=temp;
                    int now_inversion_count=getInversionCount(temp_data);
                    if(orial_inversion_count>now_inversion_count)
                    {
                        orial_inversion_count=now_inversion_count;
                    }
                    temp_data=data.clone();
                }
            }
            System.out.println(orial_inversion_count);
        }
        
    }
}