文章标题

Making the Grade
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 4732 Accepted: 2244
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

/*
题意：改动一些数据，使之非严格递增 (1 2 3 3 4 )(总的趋势是增，可以有相同数)

思路：dp[i][j] 代表前i个数改成满足题意，并且第i个数为第j小的最小消耗值
那么dp[i][j]=min(dp[i-1][k]+a[i]-b[j]) (b[j])代表排序后的第j小
其中1<=k<=j;然后发现可以用滚动数组
*/

include

include

include

include

include

include

include

include

include

include

include

typedef __int64 ll;

define L(x) (x<<1)

define R(x) (x<<1|1)

define MID(x,y) ((x+y)>>1)

using namespace std;

define INF 0x3f3f3f3f

define N 2005

ll dp[2][N];

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

int main()
{
int i,j;
while(~scanf(“%d”,&n))
{
for(i=1;i<=n;i++)
{
scanf(“%d”,&a[i]);
b[i]=a[i];
}
sort(b+1,b+n+1);
for(i=1;i<=n;i++)
dp[0][i]=fabs(a[1]-b[i]);
int now=0;
for(i=2;i<=n;i++)
{
__int64 temp=dp[now][1]; //temp记录dp[i-1][1~j]的最小值
for(j=1;j<=n;j++)
{
temp=min(temp,dp[now][j]);
dp[now^1][j]=temp+abs(a[i]-b[j]);
}
now=now^1;
}
__int64 ans=dp[now][1];

    for(i=2;i<=n;i++)
        if(dp[now][i]<ans) ans=dp[now][i];
    printf("%I64d\n",ans);

}

return 0;
}