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Consider swapping anyway, and set the last array to a a a array

Considering the meaning of absolute value, a i a_i ai} and b i b_i bi ， there must be one added and one subtracted

So if it's not fixed k k The value of k, the optimal case is equivalent to a , b a,b a. B array merge sort

front n n If n is large, give them a plus sign, followed by n n It is optimal to assign a minus sign to n numbers

So consider a couple ( a i , b i ) (a_i,b_i) (ai, bi), if the signs assigned to the optimal solution are different, there is no need to control Because in this case, the larger number must be given a positive sign, and the smaller number must be given a positive sign

So we just need to consider what a i , b i a_i,b_i ai, bi ， are also assigned positive (let the set of such pairs be S 1 S_1 S1) or a pair with a negative sign at the same time (let such a set be S 2 S_2 S2​)

Obviously, every time we exchange S 1 S_1 S1 # and S 2 S_2 S2 ﹤ that is, the minimum number of steps can reach the optimal solution, and it is assumed that it needs to be exchanged z z z times( z z z is S 1 S_1 S1 (number of elements in)

if k > = z k>=z k> = Z, then the optimal solution must be reached, because in the optimal solution a a There must be two in a + + +Number or two − - − sign, exchange such a pair of answers

if k < z k<z K < Z, it would be better to consider how to exchange

set up S 1 S_1 One pair in S1 ＾ is a i , b i a_i,b_i ai, bi ， where the smaller fraction is c c c the larger number is d d d

and S 2 S_2 The element in S2 is a j , b j a_j,b_j aj, bj, where the smaller fraction is e e e the larger number is f f f

So what are the two benefits before the exchange ( d − c ) + ( f − e ) (d-c)+(f-e) (d−c)+(f−e)

And because S 1 S_1 All elements in S1 must be greater than S 2 S_2 All elements in S2 (because they are at the front of the final order) n n n elements)

What are the benefits of these two pairs after exchange ( d − e ) + ( c − f ) (d-e)+(c-f) (d−e)+(c−f)

That is, there are more benefits after the exchange 2 ∗ ( c − f ) 2*(c-f) 2 * (c − f), that is, more 2 ∗ ( m i n ( a i , b i ) − m a x ( a j , b j ) ) \rm 2*(min(a_i,b_i)-max(a_j,b_j)) 2∗(min(ai​,bi​)−max(aj​,bj​))

So it's the same exchange k k k times, S 1 S_1 S1 ， we take min ⁡ ( a i , b i ) \min(a_i,b_i) min(ai, bi) is sorted from large to small, and S 2 S_2 S2 ＾ in m a x ( a j , b j ) \rm max(a_j,b_j) max(aj, bj) is sorted from small to large

In this way, it must be optimal to exchange the corresponding positions from front to back

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int maxn = 1e7+10;
int n,k,a[maxn],b[maxn],f[maxn];
typedef pair<int,int>p;
struct RR
{
	int val,id;
	bool operator < ( const RR&tmp ) const { return val>tmp.val; }
}S[maxn]; int top;
vector<int>S1,S2;
void solve()
{
	if( n==2 )
	{
		if( k&1 )	swap( a[1],a[2] );
		cout << abs( a[1]-b[1] )+abs( a[2]-b[2] );
		exit( 0 );
	}
	int ans = 0;
	for(int i=1;i<=n;i++)	ans += S[i].val;
	for(int i=1;i<=n;i++)	ans -= S[i+n].val;
	cout << ans;
	exit( 0 );
}
signed main()
{
	cin >> n >> k;
	for(int i=1;i<=n;i++)	cin >> a[i],S[++top] = {a[i],i};
	for(int i=1;i<=n;i++)	cin >> b[i],S[++top] = {b[i],i+n};
	sort( S+1,S+1+top ); 
	for(int i=1;i<=top;i++)	f[S[i].id] = (i<=n);
	for(int i=1;i<=n;i++)
	{
		if( f[i]!=f[i+n] )	continue;//Different number
		if( f[i] )	S1.push_back( min(a[i],b[i]) );//All positive signs 
		else	S2.push_back( max(a[i],b[i]) );//All negative signs 
	}
	if( k>=S1.size() )	solve();//Perfect match 
	sort( S1.begin(),S1.end(),[](int a,int b){ return a>b; } );
	sort( S2.begin(),S2.end(),[](int a,int b){ return a<b; } );
	int ans = 0;
	for(int i=1;i<=n;i++)	ans += abs( a[i]-b[i] );
	for(int i=0;i<k;i++)	ans += 2*( S1[i]-S2[i] );
	cout << ans;
}