/\/\/\/（摆动数组的应用&结构体数组标记法）

A sequence a1,a2,…,an is said to be /\/\/\/ when the following conditions are satisfied:

译：一个序列a1,a2,....an如果是/\/\/\/数列，那么会满足下列条件

For each i=1,2,…,n−2, ai=ai+2.
Exactly two different numbers appear in the sequence.
You are given a sequence v1,v2,…,vn whose length is even. We would like to make this sequence /\/\/\/ by replacing some of its elements. Find the minimum number of elements that needs to be replaced.

译：对于每个i=1,2，……，n-2,ai=a(i+2)，在每个序列中出现两个不相等数字。现有一个序列v1,v2,...,vn，它的长度是一个偶数，我们可以通过换一些数字来把这个序列变成一个/\/\/\/数组。输出所需替换的最小数目。

Constraints
2≤n≤1e5
n is even.
1≤vi≤1e5
vi is an integer.

 

输入

Input is given from Standard Input in the following format:

n
v1 v2 … vn

 

输出

Print the minimum number of elements that needs to be replaced.

 

样例输入

复制样例数据

4
3 1 3 2

样例输出

1

提示

The sequence 3,1,3,2 is not /\/\/\/, but we can make it /\/\/\/ by replacing one of its elements: for example, replace the fourth element to make it 3,1,3,1.

既然这个摆动数组里只能有两个数字，而且必须分别在奇数位和偶数位上，那么我们找出奇数位，偶数位上出现次数最多的数字后再用总数减去他们即可。但是，题目有说：序列里的两个数字必须不等，也就是说如果相等，要拿除了那个数之外第二多的数字来进行计算，那单纯用一个数组标记元素的办法可能就有些疲软，所以我们用结构体去标记数组的元素，恰好，这个题目里最大元素也就是1e5，显然数组标记法是可行的。

#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#include <algorithm>
#include <iomanip>
#include <cstring>
#include <cmath>
//#include <minmax.h>
#define DETERMINATION main
#define lldin(a) scanf("%lld",&a)
#define din(a) scanf("%d",&a)
#define reset(a,b) memset(a,b,sizeof(a))
const int INF=0x3f3f3f3f;
const int LIM=1e5+75;
using namespace std;
typedef long long ll;
/**you can feel the pulse of your destiny...
The nervous feeling fills you with determination.**/
int n;
int a[LIM];
struct nu
{
   int value,mx;
}hhhhhh[LIM],qqqqqq[LIM];
bool cmp(nu a,nu b)
{
    return a.mx>b.mx;
}
int DETERMINATION()
{
    cin>>n;
    for(int i=1; i<=n; i++)
        cin>>a[i];
    for(int i=1; i<=n; i+=2)//找出奇数位上出现次数最多的数字
        {
          hhhhhh[a[i]].mx++;
          hhhhhh[a[i]].value=a[i];
        }
    for(int i=2; i<=n; i+=2)//找出偶数位上出现次数最多的数字
        {
          qqqqqq[a[i]].mx++;
          qqqqqq[a[i]].value=a[i];
        }
    ll ans=0;
    sort(qqqqqq+1,qqqqqq+LIM+1,cmp);//通过排序找出最大值，由于只有1e5个元素，排序不会造成什么时间上的浪费。
    sort(hhhhhh+1,hhhhhh+LIM+1,cmp);
    //cout<<qqqqqq[1].value<<" "<<hhhhhh[1].value<<endl;
    if(qqqqqq[1].value==hhhhhh[1].value)//如果奇数位和偶数位都是同一个数字出现次数最多。
    {
       ans=min((n-hhhhhh[2].mx-qqqqqq[1].mx),(n-hhhhhh[1].mx-qqqqqq[2].mx));
       //那么枚举两种情况来求最小值。
    }
    else
      ans=n-hhhhhh[1].mx-qqqqqq[1].mx;//不然照常计算即可。
    cout<<ans<<endl;
    return 0;
}