Notes (note)

Notes
"Problem description"
Given a sequence of length m, the subscript number is 1~m. Each element of the sequence is 1~n.
Integer. The cost of defining a sequence is

M?1

∑|ai+1-ai|

I=1

You can now select two numbers x and Y, and change all x in sequence a to Y. X can be equal to Y.
The minimum possible cost of requesting a sequence.
"Input Format"
Enter the first line containing two integers n and M. The second line contains a m-space-delimited integer, which represents the ordinal
Column A.
"Output Format"
The output line contains an integer that represents the minimum cost of the sequence.
"Sample Input 1"
4 6
1 2 3 4 3 2
"Sample Output 1"
3
"Sample Input 2"
10 5
9 4 3) 8 8
"Sample Output 1"
6
"Sample Interpretation"
In Example 1, the optimal strategy is to change 4 to 3.

In Example 2, the optimal strategy is to change 9 to 4.
"Data size and conventions"
For 30% of data, n,m<=100.
For 60% of data, n,m≤2000.
For 100% of data, 1≤n,m≤100,000.

Violence can be over 6 points, which is 30 points.

The procedure is to find nearby elements and record contribution values.

Code out of the way

#include <cstdio>#include<iostream>#include<algorithm>#include<ios>#include<vector>using namespaceStd;typedefLong Longll;Const intN = (int) 1e5;intN, m, A[n +1];vector<int> B[n +1];intMainintargcChar*argv[]) {//freopen ("note.in", "R", stdin);//freopen ("Note.out", "w", stdout);iOS:: Sync_with_stdio (false);//let Cin run faster. (Don't use it easily)CIN >> N >>m;  for(inti =1; I <= m; ++i) Cin >>A[i];  for(inti =1; I <= m; ++i) {if(I >1&& A[i-1]! = A[i]) b[a[i-1]].push_back (A[i]); if(I < m && A[i +1]! = A[i]) B[a[i +1]].push_back (A[i]); }//find the nearby element. ll ans= 0LL, sum =0LL;  for(inti =1; I <= N; ++i) {if(!b[i].size ())Continue;        Sort (B[i].begin (), B[i].end ()); inty = b[i][b[i].size () >>1]; ll before= 0LL, after =0LL;  for(intj =0; J < B[i].size (); ++j) {before+ = ABS (I-B[i][j]); after+ = ABS (Y-B[i][j]); } ans= Max (ans, before-after), sum + =before; } cout<< sum/2-Ans << Endl;//because the two sides are added, so/2. fclose (stdin);    Fclose (stdout); return 0;}

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Notes (note)