线性DP POJ3666 Making the Grade

Making the Grade

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7347		Accepted: 3415

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, ... , A_N (1 ≤ N ≤ 2,000) describing the elevation (0 ≤ A_i ≤ 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, . ... , B_N 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| + ... + |A_N - B_N |

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: A_i

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

 

引理一下，记|Ai-B_i|的和为S，在S最小化前提下，一定存在一种构造序列B的方案，使B中的数值都在A中出现过。

然后很容易做了……

附lyd的讲解

 

对于这个引理，简单地证明一下：

可以用数学归纳法证明。

假设b[1~k-1]都在a[]中出现过；

对于b[k]，若a[k]>b[k-1]，则b[k]=a[k]解最优；

　　　　  若a[k]<b[k-1]，则一定存在t，使b[t~k]变为a[k]的最优解；

　　　　　　　　　　　　 或b[k]=b[k-1]；

特殊地，对于1号位置，b[1]=a[1]必定为最优解。

综上，引理显然得证。

接下来进行dp

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n;
int a[2010],num[2010],f[2010][2010];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]),num[i]=a[i];
    sort(num+1,num+n+1);
    memset(f,0x3f3f3f3f,sizeof(f));
    f[0][0]=0;
    int tmp=0;
    for(int i=1;i<=n;i++){
        tmp=f[i-1][0];
        for(int j=1;j<=n;j++){
            tmp=min(tmp,f[i-1][j]);
            f[i][j]=abs(a[i]-num[j])+tmp;    
        }
    }
    int ans=0x3f3f3f3f;
    for(int i=1;i<=n;i++)
        ans=min(ans,f[n][i]);
    printf("%d
",ans);
    return 0;
}