https://code.google.com/codejam/contest/11304486/dashboard

Simple problem.
Just sort all numbers, and for each i<j, there are 2^(j-i) subsets with min-a[i] and max-a[j].

For each point (a, b), make two diagonal lines y=x+b-a, y=-x+a+b.
Then, the final point must be an intersection of two lines.

For small input, just check all intersections to calculate the min-value. O(n^3).

For large input, we need to calculate in O(n^2).
For any diagonal line L, there are n intersections on the line.
Let's assume points Pi=(ai, bi), the intersection of L and anti-diagonal line of Pi is Qi=(ci, di).
Then, for X=(x, y):

dis(X, Pi)=max(|x-ai|, |y-bi|)
dis(X, Pi)=dis(X, Qi)+dis(Qi, Pi)

As all Qi can be calculated by Pi, we just need to calculate

sum = sigma_(i,1,n) {dis(X,Qi)}
= sigma_(i,1,n) {w[i]*|x-ci|}

So, the problem is, there are N points on a line, (c[1], c[2], ..., c[n]), calculate the minimal value of s(x) = sigma_(i,1,n) {w[i]*|x-c[i]|.
Let x[0]=c[0], s[0]=s(x[0]), q[0]=-sigma {w[i]}.
Then for 1<=j<=n,

x[j]=c[j]
dx=x[j]-x[j-1]
s[j]=s[j-1]+dx*q[j]
q[j]+=2*w[j]

Finally, calculate the minimal value of s[i].

DP problem.
For each position (i, j) and each 1<=v<=k, dp[i][j][v] means the largest v-Christmas tree with top (i,j). We just need to calculate the max value of dp[i][j][k].
To do this, just calculate the height of each point, then dp.