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要点：DP+辅助数组优化

题解：

由题意易得出DP方程

f[i][j] = min(f[i-1][k]+|j-k|*c+(a[i]-j)*(a[i]-j))

时间复杂度为O(N*100*100)

摘出与k无关项得

f[i][j] = min(f[i-1][k]+|j-k|*c) + (a[i]-j)*(a[i]-j)

记P = min(f[i-1][k]+|j-k|*c) , Q = (a[i]-j)*(a[i]-j)

则f[i][j] = P + Q

易知

P = min(A,B),其中

A = min(f[i-1][k]+(j-k)*c) (k<=j)

B = min(f[i-1][k]+(k-j)*c) (k>j)

A = min(f[i-1][k]-k*c) + j*c

B = min(f[i-1][k]+k*c) - j*c

记C[X][i] = min(f[X][k] - k*c)  k∈[1,i]

记D[X][i] = min(f[X][k] + k*c)  k∈[i,n]

则A = C[i-1][j] + j*c

则B = D[i-1][j+1] - j*c

显然C、D在任何时刻只需保存X=i-1一行的值

注意高度只能增高，所以h[i]∈[a[i],100]

利用辅助数组优化后，时间复杂度降为O(N*100)

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>//
#include<algorithm>
#include<iostream>
using namespace std;

#define rep(i,a,b) for(int i=a; i<=b; i++)
#define dow(i,a,b) for(int i=a; i>=b; i--)
#define tab(i,x) for(int i=head[x]; i!=-1; i=e[i].next)
#define cls(a,x) memset(a, x, sizeof a);
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 100010, H = 110;
int n,c;
int h;
int a
;
int P,Q,A,B,C[H],D[H];
int f
[H];
int main() {
scanf("%d%d",&n,&c);
rep(i,1,n) scanf("%d",&a[i]);
h=a[1];
rep(i,1,n) h=max(h,a[i]);
h=min(h,100);

cls(f,0x3f);

rep(i,1,n) {
if(i==1) {
rep(j,a[1],h) f[1][j] = (j-a[1])*(j-a[1]);
}
else {
rep(j,a[i],h) {
Q = (j-a[i])*(j-a[i]);
A = C[j] + j*c;
B = D[j+1] - j*c;
P = min(A,B);
f[i][j] = P + Q;
}
}

C[0] = D[h+1] = INF;
rep(j,1,h) C[j] = min(C[j-1], f[i][j] - j*c);
dow(j,h,1) D[j] = min(D[j+1], f[i][j] + j*c);
}

int ans=INF;
rep(i,1,h) ans = min(ans, f
[i]);
printf("%d\n",ans);
return 0;
}