CF-13C. Sequence(DP)

C. Sequence

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

Little Petya likes to play very much. And most of all he likes to play the following game:

He is given a sequence of N integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.

The sequence a is called non-decreasing if a₁ ≤ a₂ ≤ ... ≤ a_N holds, where N is the length of the sequence.

Input

The first line of the input contains single integer N (1 ≤ N ≤ 5000) — the length of the initial sequence. The following N lines contain one integer each — elements of the sequence. These numbers do not exceed 10⁹ by absolute value.

Output

Output one integer — minimum number of steps required to achieve the goal.

Sample test(s)

Input

5
3 2 -1 2 11

Output

4

Input

5
2 1 1 1 1

Output

1

 

思路：ff[]为原始序列，f[i]为原始序列的排序。dp[x][y]为把前x个数以j为标杆（0<=j<=y）变成非递减的最优值。

          答案就是dp[m-1][m-1];

        方程就是

         dp[x][y]=min(dp[x][y-1],dp[x-1][y]+abs(ff[x]-f[y]))

         优化空间。dp[j]为以第0-j个为标杆的把前i个数变成非递减的最小费用的最优值

       dp[x]=min(dp[x-1],dp[x]+abs(ff[i]-f[x]));

盲点：为什么是用原始序列的排序为标杆得到的值就是最优的呢？以这些值以外的值为标杆就不可能得到更优的值吗？求解。

 

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int mm=5004;
long long f[mm],ff[mm],dp[mm];
int main()
{
  int m;
  while(cin>>m)
  {
    for(int i=0;i<m;i++)
    {
      cin>>f[i];ff[i]=f[i];
    }
    sort(f,f+m);///f为排完序的数列
    memset(dp,0,sizeof(dp));
    ///dp[j]为以第j个为标杆的把前i个数变成非递减的最小费用
    for(int i=0;i<m;i++)
      for(int j=0;j<m;j++)
    {
      dp[j]+=abs(ff[i]-f[j]);///以f[j]为标杆
      dp[j]=j&&dp[j]>dp[j-1]?dp[j-1]:dp[j];
      ///更新值，dp[j]是以0-j为标杆中把前i个变成非递减的最小费用最优值
    }
    cout<<dp[m-1]<<"\n";
  }
}