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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 逆序數的概念：在一個排列中，如果一對數的前後位置與大小順序相反，即前面的數大於後面的數，那末它們就稱為一個逆序。
一個排列中逆序的總數就稱為這個排列的逆序數。
思路：先暴力求出開始時的逆序數，然後會有一個性質：每次把末尾的數掉到序列前面時，減少的逆序對數為n-1-a[i] ，增加的逆序對數為a[i] ，所以求出其他情況時的逆序數。

 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 }

1394-Minimum Inversion Number