Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20162 Accepted Submission(s):
12110

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

Author

CHEN, Gaoli

逆序数的概念：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，那末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。
思路：先暴力求出开始时的逆序数，然后会有一个性质：每次把末尾的数掉到序列前面时，减少的逆序对数为n-1-a[i] ，增加的逆序对数为a[i] ，所以求出其他情况时的逆序数。

 #include<cstdio>

 int t[];

 int n,ans,sum;

 int main()

 {

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

     {

         sum = ;

         for (int i=; i<=n; ++i)

             scanf("%d",&t[i]);

         for (int i=; i<n; ++i)

             for (int j=i+; j<=n; ++j)

                 if (t[i]>t[j]) sum++;

         ans = sum;

         for (int i=n; i>=; --i)

         {

             sum -= n--t[i];

             sum += t[i];

             if (sum < ans) ans = sum;

         }

         printf("%d\n",ans);

     }

     return ;

 }

巴特西

1394-Minimum Inversion Number

Minimum Inversion Number

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