S - Making the Grade 【滚动数组+线性离散化】

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：求变成单调不上升 单调不递减的最小花费
既有滚动数组又有线性离散化 很不错的题~

感觉C++库里面的函数，有点有趣= =

#include 
#include 
#include 
#include 
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 1e5 + 10;
int n, a[2005], b[2005], dp[2005];
int solve() 
{
    memset(dp, 0, sizeof dp);
    for (int i = 1; i <= n; ++i)
    {
        int preMin = INF;
        for (int j = 1; j <= n; ++j)
        {
            preMin = min(preMin, dp[j]);
            dp[j] = preMin + abs(a[i] - b[j]);
        }
    }
    return *min_element(dp + 1, dp + 1 + n);//最大值函数
}

int main() 
{
    while (cin >> n)
    {
        for (int i = 1; i <= n; ++i)
        {
            cin >> a[i], b[i] = a[i];
        }

        sort(b + 1, b + 1 + n);
        int ans = INF;
        ans = min(ans, solve());
        reverse(b + 1, b + 1 + n);//反转
        ans = min(ans, solve());
        cout << ans << endl;
    }
    return 0;
}
/*
7
1
3
2
4
5
3
9
*/