2020 Niuke Summer Multi-School Training Camp (third session) E. Two Matchings structure + dp

Topic link: E.Two Matchings

Title

Let you construct two sequences p, q of length n. The two sequences have different bits, and satisfy $p_i\ne iand{p}_{p_i}=i$ . Given you a sequence a, you can calculate a value$({\sum }_{i=1}^{n}abs(a_i-a_{p_i}\left)\right)/2$ Find the smallest sum of calculated values ​​among p and q that meet the conditions.

answer

This question first analyzes the sequence of p and q. The question gives $p_i\ne iand{p}_{p_i}=i$ , in fact, the sequence of p, q is equivalent to one, two, three...n swapping positions once. Since n is an even number, there will be no such thing as a position swapping multiple times.
If required$({\sum }_{i=1}^{n}abs(a_i-a_{p_i}\left)\right)/2\; is\; the$ smallest, then we can easily determine a sequence first, that is, after sequence a is sorted, every two adjacent ones can be exchanged.

So how to determine the second smallest value?
If n=4, then the sequence of the next smallest value is: 4 1 3 2

If n is 6, then the sequence of the next smallest value is: 6 3 2 5 4 1

At first I was thinking why not when n=6: 3 1 4 2 6 5, and finally found that this is the same as the sequence we constructed for the first time: 2 1 4 3 6 5 The last two digits are the same, which is not suitable for the purpose.

Now that we know the minimum value of n=4 and n=6, do we still need to enumerate the minimum value of n=8,10...?
Of course not, $gcd(4,6)=2.\; According\; to$ Pei Shu's theorem in number theory,$4k_1+6k_2=gcd(4,6)=2$ is solvable, then it means thatthe sum of multiples of 4 and multiples of 6 must be a multiple of 2.
The implication is that the minimum value of 4 and 6 is known, and the remaining n (even number) can be solved by recursion.

Let dp[i]: the minimum value of the sum of the previous i numbers.
Initialization:
dp[2]=inf
dp[4]=a[4]+a[3]-a[2]-a[1]
dp[6]=a[6]+a[5]-a[4 ]+a[3]-a[2]-a[1]
Then the state transition equation is:
dp[n]=min(dp[i-4]+a[i]+a[i-1]-a[ i-2]-a[i-3],
dp[i-6]+a[i]+a[i-1]-a[i-2]+a[i-3]-a[i-2 ]-a[i-1])

Code

#include<iostream>
#include <sstream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
//extern "C"{void *__dso_handle=0;}
typedef long long ll;
typedef long double ld;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define lowbit(x) x&-x
//#define int long long

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const ll inf=1e18;
const int maxn=2e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

ll a[maxn];
ll dp[maxn];
int main()
{
    
    
	int t;
	scanf("%d",&t);
	while(t--)
	{
    
    
		int n; scanf("%d",&n);
		for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
		sort(a+1,a+1+n);
		ll sum=0;
		for(int i=2;i<=n;i+=2) sum+=a[i]-a[i-1];
		dp[2]=inf;
		dp[4]=a[4]+a[3]-a[2]-a[1];
		dp[6]=a[6]+a[5]+a[3]-a[4]-a[2]-a[1];
		for(int i=8;i<=n;i+=2)
		{
    
    
			dp[i]=min(dp[i-4]+(a[i]+a[i-1]-a[i-2]-a[i-3]),
			(dp[i-6]+a[i]+a[i-1]-a[i-2]+a[i-3]-a[i-4]-a[i-5]));
		}
		sum+=dp[n];
		printf("%lld\n",sum);
	}
}