HDU 1394 Minimum Inversion Number

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本文介绍了一种算法，用于解决给定序列通过循环移动首位元素产生的所有序列中逆序对数量最小的问题。通过计算原始序列的逆序对数量，并在每次移动后更新逆序对数量，最终找到逆序对数量最小的序列。

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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

题目大意
给你一段序列a1,a2,a3,a4,a5,a6,a7,,然后这段序列可以一直调动第一个数字到最后，例如：
a2,a3,a4,a5,a6,a7,a1,…………直到循环了一遍。然后从这么多的序列中求出逆序列最小的一组。注意，序列中的数是0-n-1。
先求出原序列的逆序对。
原序列每个数加1。
将第一个数移到列尾，就会提供n-a[i]个逆序对，减a[i]-1个逆序对。
模拟即可。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,a[5005],ans,tmp,c[5005];
int lowbit(int x)
{
    return x&(-x);
}
int getsum(int pos)
{
    int sum=0;
    while(pos>0)
    {
        sum+=c[pos];
        pos-=lowbit(pos);
    }
    return sum;
}
void add(int pos)
{
    while(pos<=n)
    {
        ++c[pos];
        pos+=lowbit(pos);
    }
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        memset(c,0,sizeof(c));
        tmp=0,ans=1e9+7;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",a+i);
            ++a[i];
        }
        for(int i=1;i<=n;i++)
        {
            tmp+=i-1-getsum(a[i]);
            add(a[i]);
        }
        ans=min(tmp,ans);
        for(int i=1;i<=n-1;i++)
        {
            tmp+=n-2*a[i]+1;
            ans=min(ans,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}