归并求逆序对数

先上模板，复杂度为O（n）

void merge_sort(int *a, int l, int r, int *t){
  if(r - l > 1){
    int mid = l + (r-l)/2;
    int p = l, q = mid, i = l;
    merge_sort(a, l, mid, t);
    merge_sort(a, mid, r, t);

    while(p < mid || q < r){
      if(q >= r || (p < mid && a[p] <= a[q]))
        t[i++] = a[p++];
      else {
        t[i++] = a[q++];
        cnt += mid-p;
      }
    }
    for(i = l; i < r; i ++)
      a[i] = t[i];
  }
}

逆序数的定义：对于i < j 有 a[i] > a[j]。首先brute force，是对于每个a[i]，求逆序对数。

例题：

zoj1484

求序列

a1, a2,.....an

a2,  .....an,a1

....

an, a1,.......an-1

逆序数的最小值。

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

最后

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