S - Making the Grade POJ - 3666

A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ would like to add and remove dirt from the road so that it becomes one monotonic slope (either sloping up or down).

You are given N integers A1, ... , AN (1 ≤ N ≤ 2,000) describing the elevation (0 ≤ Ai ≤ 1,000,000,000) at each of N equally-spaced positions along the road, starting at the first field and ending at the other. FJ would like to adjust these elevations to a new sequence B1, . ... , BN that is either nonincreasing or nondecreasing. Since it costs the same amount of money to add or remove dirt at any position along the road, the total cost of modifying the road is

| A 1 - B 1| + | A 2 - B 2| + ... + | AN - BN |

Please compute the minimum cost of grading his road so it becomes a continuous slope. FJ happily informs you that signed 32-bit integers can certainly be used to compute the answer.

Input

* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single integer elevation: Ai

Output

* Line 1: A single integer that is the minimum cost for FJ to grade his dirt road so it becomes nonincreasing or nondecreasing in elevation.

Sample Input

7
1
3
2
4
5
3
9

Sample Output

3

题意：把序列A变成序列B使得B非单调递增或者非单调递减，花费为|A[i]-B[i]|，求最小花费。

dp[i][j]表示前i - 1个数已经有序的情况下把第i个数改变为第j小的数的最小花费。

dp[i][j] = min(dp[i - 1][k] + fabs(a[i] - b[j]))  (k <= j)

现在我还是有点不能理解，以后还要多想想。

#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 2020;
const long long inf = 1e15;
long long dp[maxn][maxn],a[maxn],b[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;i++)
    {
        scanf("%lld",&a[i]);
        b[i] = a[i];
    }

    sort(b + 1,b + 1 + n);

    for(int i = 1;i <= n;i++)
        dp[1][i] = fabs(a[1] - b[i]);

    for(int i = 2;i <= n;i++)
    {
        long long minn = dp[i - 1][1];
        for(int j = 1;j <= n;j++)
        {
            minn = min(dp[i - 1][j],minn);
            dp[i][j] = minn + fabs(a[i] - b[j]);
        }
    }

    long long ans = inf;
    for(int i = 1;i <= n;i++)
        ans = min(ans,dp[n][i]);

    printf("%lld\n",ans);

    return 0;
}