整理篇

Making the Grade

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5797		Accepted: 2714

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

/*
解题思路整理--题意：花费最少代价使序列单调不下降或不上升 
f[i,j]表示第i段路升高到是s[j]，abs(h[i]-s[j])为原高度与ans对应高度的差 
于是f[i,j]=min(f[i-1,k]+abs(h[i]-s[j]))（k<=j)
o(n3)要tle的节奏。
那么预处理--f[i-1,k]：
开一个数组g[i,j]表示前i段路中升高到s[j]的最小花费,
g[i,j]=min(g[i,j-1],f[i,j])。
最终：f[i,j]=g[i-1,j]+abs[a[i]-c[j]]; o(n2)
具体实现--sort s 正排一遍，倒排一遍 dp2遍--AC 
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define N 2010
using namespace std;
int s[N],h[N],n,res=2137483648;
int f[N][N],g[N][N];
int cmp(int a,int b){
    return a>b;
}
void dp(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            f[i][j]=g[i-1][j]+abs(h[i]-s[j]);
            if(j==1) g[i][j]=f[i][j];
            else g[i][j]=min(g[i][j-1],f[i][j]);
        }
    }
    for(int i=1;i<=n;i++){
        res=min(res,f[n][i]);
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",h+i);s[i]=h[i];
    }
    sort(s+1,s+n+1);dp();
    sort(s+1,s+n+1,cmp);dp();
    printf("%d
",res);
    return 0;
}