USACO Making the Grade
洛谷 P2893 [USACO08FEB]修路Making the Grade
https://www.luogu.org/problemnew/show/P2893
JDOJ 2566: USACO 2008 Feb Gold 1.Making the Grade
https://neooj.com:8082/oldoj/problem.php?id=2566
POJ Making the Grade
http://poj.org/problem?id=3666
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 A_1, . . . , 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 B_1, . . . , 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
|A_1 - B_1| + |A_2 - B_2| + ... + |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
1
3
2
4
5
3
9
Sample Output
HINT
OUTPUT DETAILS:
By changing the first 3 to 2 and the second 3 to 5 for a total cost of
|2-3|+|5-3| = 3 we get the nondecreasing sequence 1,2,2,4,5,5,9.
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,ans,a[],t[],b[];
int f[][],minf[][];
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
scanf("%d",&n);
for(int i=;i<=n;i++)
{
scanf("%d",&a[i]);
t[i]=a[i];
}
sort(t+,t+n+);
int now=-;
for(int i=;i<=n;i++)
if(now!=t[i])
b[++m]=t[i],now=t[i];
for(int i=;i<=n;i++)
for(int j=;j<=m;j++)
{
f[i][j]=minf[i-][j]+abs(a[i]-b[j]);
if(j==)
minf[i][j]=f[i][j];
else
minf[i][j]=min(minf[i][j-],f[i][j]);
}
ans=minf[n][m];
memset(f,,sizeof(f));
memset(minf,,sizeof(minf));
sort(b+,b+m+,cmp);
for(int i=;i<=n;i++)
for(int j=;j<=m;j++)
{
f[i][j]=minf[i-][j]+abs(a[i]-b[j]);
if(j==)
minf[i][j]=f[i][j];
else
minf[i][j]=min(minf[i][j-],f[i][j]);
}
ans=min(ans,minf[n][m]);
printf("%d",ans);
return ;
}
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