文章标题includeincludeincludeincludeincludeincludeincludeincludeincludeincludeincludedefine L(x) (x<<1)define R(x) (x<<1|1)define MID(x,y) ((x+y)>>1)define INF 0x3f3f3f3fdefine N 2005

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 

9 

Sample Output

Source

/* 

题意：修改一些数据，使之非严格递增 (1 2 3 3 4 )(总的趋势是增，能够有同样数)

思路：dp[i][j] 代表前i个数改成满足题意，而且第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 

当中1<=k<=j;然后发现能够用滚动数组 

*/

typedef __int64 ll;

using namespace std;

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); 

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];

return 0; 

}

本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5261599.html，如需转载请自行联系原作者