Service Stations on a Line: Consecutive Groups and Medians¶

Turn placement into grouping¶

Sort the coordinates as \(x_1\le\cdots\le x_N\) . In station order, each station can serve a consecutive interval of homes; distance ties can also be assigned consecutively. For any fixed group, a median minimizes the sum of absolute distances. Thus, an optimum exists with every station at a home coordinate.

We may use exactly \(M\) nonempty groups. If a station serves no home, \(M\le N\) lets us split a group containing at least two homes and add a station without increasing the cost. Duplicate coordinates do not affect this argument.

Let \(P_i=\sum_{j=1}^ix_j\) and \(P_0=0\) . The minimum cost of serving \([l,r]\) with one station is

\[ C(l,r)=x_q(q-l+1)-(P_q-P_{l-1})+(P_r-P_q)-x_q(r-q),\qquad q=\lfloor(l+r)/2\rfloor. \]

Prefix sums evaluate it in constant time.

Dynamic programming and monotone decisions¶

Let \(D_t(r)\) be the minimum cost of splitting the first \(r\) homes into \(t\) groups. Set \(D_0(0)=0\) and every other infeasible state to infinity. Then

\[ D_t(r)=\min_{t-1\le j<r}\{D_{t-1}(j)+C(j+1,r)\},\qquad r\ge t. \]

Direct evaluation costs \(O(MN^2)\) . The interval cost is Monge: for \(a\le b\le c\le d\) ,

\[ C(a,c)+C(b,d)\le C(a,d)+C(b,c). \]

To prove this, consider the increment

\[ C(l,r+1)-C(l,r)=x_{r+1}-x_{\lfloor(l+r+1)/2\rfloor}. \]

For fixed \(r\) , this increment cannot increase as \(l\) increases. Summing the increments from \(c\) to \(d\) proves the inequality. Adding \(D_{t-1}(j)\) to candidate column \(j\) preserves it, so the smallest optimal split is nondecreasing with \(r\) . If two optima cross, the inequality allows exchanging them without increasing the cost; choosing the smallest split on ties rules out a strict inversion.

Divide and conquer within one layer¶

Suppose targets \([L,R]\) have optimal splits in \([A,B]\) . Evaluate the middle target \(r\) by trying \([A,\min(B,r-1)]\) . If its best split is \(j\) , recurse on the left targets with upper bound \(j\) , and on the right targets with lower bound \(j\) . Each layer starts with targets \([t,N]\) and candidates \([t-1,N-1]\) . Update the best split only for a strictly smaller value to keep the smallest-index tie rule.

The recurrence considers every possible final group. Monotonicity ensures that the restricted recursive searches retain an optimum. Therefore, the answer is \(D_M(N)\) .

Implementation, complexity and comparison¶

Each recursion depth scans \(O(N)\) candidates in total, and there are \(O(\log N)\) depths per layer. Including sorting, time is \(O(N\log N+MN\log N)\) and extra space is \(O(N)\) with two rolling layers. Costs are at most \(N\cdot10^8\) ; use 64-bit integers and \(2^{60}\) as infinity.

std.cpp is the reference. The setter's method tracks the rightmost station and precomputes the cost between consecutive stations. Its complete endpoint-placement explanation and alternative.cpp are provided. It takes \(O(N^2+MN\log N)\) time and \(O(N^2)\) space. Median grouping removes the quadratic table. Both support the full constraints; see the measurements.

Setter contribution: endpoint states, interval costs and divide-and-conquer optimization. GPT-6 applies classical median grouping and Monge optimization, and supplies proofs, bilingual explanations and standard C++17 programs for both methods. Small cases use naive interval DP; formal cases use an independent SMAWK row-minimum algorithm.