Minimum Inversion Number

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1   Accepted Submission(s) : 1

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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



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先百度逆序数：在一个排列中，如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数。逆序数为​​偶数​​的排列称为偶排列；逆序数为奇数的排列称为奇排列。如2431中，21，43，41，31是逆序，逆序数是4，为偶排列。

对于这个题求把第一个数放到最后一个数的最小逆序数，对于原序列而言，如果把第一个数放到最后一个数，逆序列增加n-num[0]+1,逆序列减少

num[0].

所以这道题：

#include <stdio.h>
int main()
{
  int num[5005],n;
  while(scanf("%d",&n)!=EOF)
  {
    for(int i=0;i<n;i++)
    scanf("%d",&num[i]);
    int sum=0,temp;
    for(int i=0;i<n;i++)
    for(int j=i+1;j<n;j++)
    if(num[i]>num[j]) sum++;
    temp=sum;
    for(int i=n-1;i>=0;i--)
    {
      temp-=n-1-num[i];
      temp+=num[i];
      if(temp<sum)
      sum=temp;
    }
    printf("%d\n",sum);
  }
  return 0;
}