HDU 1394 Minimum Inversion Number

题目链接：HDU 1394 Minimum Inversion Number

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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

题意：

一个由0..n-1组成的序列，每次可以把队首的元素移到队尾，求形成的n个序列中最小逆序对数目

AC代码：

#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <algorithm>#include <iostream>#include <locale>#include <vector>#include <string>#include <iomanip>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <functional>#define maxn 100007using namespace std;int main(){    int n;    int a[5005];    while(~scanf("%d",&n))    {        long long minn=0;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            for(int j=0;j<i;j++)            {                if(a[j]>a[i])                    minn++;            }        }        long long ans=minn;        for(int i=0;i<n;i++)        {            ans=ans+n-2*a[i]-1;            if(ans<minn)                minn=ans;        }        printf("%lld\n",minn);    }    return 0;}