# Making the Grade (POJ 3666)

At first, when we add height or remove height at a point, we should always make it to a existing height in current grade. There is no need to doing more or less. Based on this we can give a DP solution.

Let's say the original array is , at first we sort the current heights into an array . Let be that the minimum cost with the first items in and we make the slope ends in a height of . Then we have our DP formular:

We can same the minimum during iteration to reduce running time. With this solution we need not care if the slope is going up or down.

import java.io.*;
import java.util.*;

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int N = in.nextInt();
int[] elev = new int[N];
int[] elevsort = new int[N];
for (int i = 0; i < N; i++) {
elev[i] = in.nextInt();
elevsort[i] = elev[i];
}
Arrays.sort(elevsort);

int[] dp = new int[N];
int[] min = new int[N];
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
dp[j] = min[j] + Math.abs(elev[i] - elevsort[j]);
}
min[0] = dp[0];
for (int j = 1; j < N; j++) {
min[j] = Math.min(min[j - 1], dp[j]);
}
}

int res = Integer.MAX_VALUE;
for (int n : min) {
res = Math.min(res, n);
}

System.out.println(res);
}
}