[POJ] 3666 Making the Grade

Making the Grade
Time Limit: 1000MS      Memory Limit: 65536K
Total Submissions: 8732     Accepted: 4088
Description

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

|A1 - B1| + |A2 - B2| + ... + |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
Source

USACO 2008 February Gold

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经过证明可以知道选出的数字一定在a中出现过。
（待补充）

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先考虑单调不降的情况
f[i][j]为前i位，最后一位选取的数字是j的情况。
j也就是最大值了。

f[i][j]=min(f[i-1][k])+abs(a[i]-j] 其中1<=k<=j

就像LICS一样，转移可以由O(n)压为O(1)

简单地讲，就是不需要真的遍历[1,j]的最小值，它是由前往后转移的，维护决策集合的最小值，加入新值的时候判断是否可以更新决策集合的最小值mn。

不升同理，但是这题数据有问题，不降就可以过了。

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//Stay foolish,stay hungry,stay young,stay simple
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

const int MAXN=2002;

int n;
int a[MAXN],b[MAXN];

int f[MAXN][MAXN];

int main(){
    cin.sync_with_stdio(false);
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i],b[i]=a[i];
    sort(b+1,b+1+n);
    for(int i=1;i<=n;i++){
        int mn=f[i-1][1];
        for(int j=1;j<=n;j++){
            mn=min(mn,f[i-1][j]);
            f[i][j]=mn+abs(a[i]-b[j]);
        }
    }
    int ans=1<<30;
    for(int i=1;i<=n;i++) ans=min(ans,f[n][i]);
    cout<<ans<<endl;
    return 0;
}