Eight clever but useless training

https://www.luogu.org/problem/P2507

analysis:

This question will not be the first to see what advanced algorithms, certainly greedy ah ah dp

First consider the ai may be equal to BI , directly to sort

But ai! = Bi's , how do?

Read the title! ! !

God gave you shut a door will give you open a window de

arrays a and b are unequal elements within the array

So it is necessary to exchange
At this time, the smaller the exchange range, the absolute value and the smaller

Our greedy to think, because each number at most in a, b each in appears once ,

So when a particular ai = bi , then ai! = B (i-1 ), ai! = B (i + 1)

Because it is an absolute value, you do not know ai-b (i-1) and b (i + 1) -ai who more

So to take a minimum

a1-b1
a1-b2, a2-b1 (twenty-two exchange)
a1-b2, a2-b3, (exchange between three) a3-b1
a1-b3, a2-b1, (exchange between three) a3-b2

Then you will want to ask

So as not to repeat it?

Consider it a (i-1), ai , a (i + 1) becomes AI, A (I +. 1), A (I + 2) , so they push past the linear DP , is not repeated

If you're still not sure about it, see the code will understand

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int N=100005;
const long long M=1e13;
int a[N],b[N],n;
long long f[N],ans;

long long Min(long long a,long long b)
{
if (a<b) return a;
return b;
}
long long js(int l,int r)
{
if (a[l]==b[r]) return M;
return abs(a[l]-b[r]);
}
int main()
{
scanf("%d",&n);
for (int i=1; i<=n; i++) scanf("%d%d",&a[i],&b[i]);
sort(a+1,a+n+1);
sort(b+1,b+n+1);
if (n==1&&a[1]==b[1]) printf("-1\n");
else
{
    f[0]=0;
    for (int i=1; i<=n; i++)
    {
        ans=M;
        if (i>=1) ans=Min(ans,f[i-1]+js(i,i)); 
        if (i>=2) ans=Min(ans,f[i-2]+js(i-1,i)+js(i,i-1));
        if (i>=3) 
        {
            ans=Min(ans,f[i-3]+js(i-2,i-1)+js(i-1,i)+js(i,i-2));
            ans=Min(ans,f[i-3]+js(i-2,i)+js(i-1,i-2)+js(i,i-1));
        } 
        f[i]=ans;
    }
    printf("%lld\n",f[n]);
}
return 0;
}

Eight clever but useless training

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