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
有一个n个整数的排列,这n个整数就是0,1,2,3...n-1这n个数( 但不一定按这个顺序给出)。现在先计算一下初始排列的逆序数,然后把第一个元素a1放到an后面去,形成新排列a2 a3 a4...an a1,然后再求这个排列的逆序数。继续执行类似操作(一共要执行n-1次)直到产生排列an a1 a2...an-1为止。计算上述所有排列的逆序数,输出最小逆序数。
假设树状数组为c[i],它所要加速的数组为a[i],输入的为b[i],我们先只考虑a[i]和b[i],朴素的想法是每输入一个b[i],就判断从0<=j<=i-1,如果b[j]>b[i],那么ans++(当然,这道题完全可以这么水过去)。那么现在,每输入一个b[i],我就用a[b[i]+1]++去标记总共输入多少次了(因为输入的数据是0~n-1,所以输入的b[i]就相当于a[i]的下标i,而且有0,树状数组无法对下标为0进行操作,所以要a[b[i]+1]),那么现在很明显了,对于一个b[i],要想查询它前面有多少大于它的,只需将a[b[i]+1]到a[n]加起来,也就是求一段数组的和,那么树状数组就上场加速了。
AC代码: