Minimum Inversion Number（树状数组）

Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21282 Accepted Submission(s): 12760

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

//这道题一开始题意理解错了 后来和同学商量了一下 是我理解有问题
题意是这样的有n个序列
a1a2…….an
a2a3……ana1
..
.
.
.
.
ana1a2…..an-1
问这里面的那个的逆序数最小
这道题就是我之前整理的逆序数的一个变形
其实我们可以这样想 就是我们先求出来第一个的逆序数
那么剩下的逆序数都是有其上一个状态得过来的
如果我们要把a[1]移动到最后面 那么新的序列的逆序数 就是之前的逆序数 + 比a[1]大的个数 (由于a[1]移动到最后面 所以前面所有比a[1]大的都是a[1]的逆序数) - 所有比a[1]小的数(因为 a[1]在没移动之前 是位于序列第一个元素 所以 比a[1]小的数都会差生一个逆序数 这样一来一共会产生a[i] - 1个) 所以通项就是
ans += (n - a[i]) - (a[i] -1 );

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int sum[5007],a[5007];int n;int lowbit(int x){    return x&(-x);}void change(int x,int t){    while(x<=n)    {        sum[x] +=t;        x +=lowbit(x);    }}long long getsum(int x){    long long ans = 0;    while(x>0)    {        ans +=sum[x];        x -=lowbit(x);    }    return ans;}int main(){    while(~scanf("%d",&n))    {        int ans =0;        memset(sum,0,sizeof(sum));//千万别忘记初始化 我之前忘记初始化 一直WA 加上了才AC        for(int i = 1;i<=n;i++)        {            scanf("%d",&a[i]);            a[i]++;            change(a[i],1);            ans +=(i  - getsum(a[i]));        }        long long mn = ans;        for(int i = 1;i<n;i++)        {            ans +=(n - a[i] - (a[i] - 1));            if(mn >ans) mn = ans;        }        printf("%lld\n",mn);    }    return 0;}