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

16

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。

 1 #include<cstdio>
 2
 3 int t[5010];
 4 int n,ans,sum;
 5 int main()
 6 {
 7     while (scanf("%d",&n)!=EOF)
 8     {
 9         sum = 0;
10         for (int i=1; i<=n; ++i)
11             scanf("%d",&t[i]);
12         for (int i=1; i<n; ++i)
13             for (int j=i+1; j<=n; ++j)
14                 if (t[i]>t[j]) sum++;
15         ans = sum;
16         for (int i=n; i>=1; --i)
17         {
18             sum -= n-1-t[i];
19             sum += t[i];
20             if (sum < ans) ans = sum;
21         }
22         printf("%d\n",ans);
23     }
24     return 0;
25 }