在给定约束下构造满足指定路径数的有向无环图是否可行？

Absolutely yes! You can always construct such a DAG that meets all the given constraints (n ≤ 300, m ≤ 400, no self-loops, acyclic) for any x ≤ 10^18. Here's a concrete, easy-to-follow construction method:

Core Idea #

We use the binary representation of x—since every integer can be written as a sum of distinct powers of 2. We’ll build a DAG that can generate each power of 2 as a path count, then combine these components to sum up exactly to x.

Step-by-Step Construction #

1. Node Setup

   • Start with node 1 (the source) and node n (the sink).
   • Create 61 intermediate nodes v₀, v₁, ..., v₆₀ (since 2⁶⁰ ≈ 1.15×10¹⁸, which covers all x ≤ 10¹⁸).
   • Create 60 auxiliary nodes u₀, u₁, ..., u₅₉ to enable path count doubling.

2. Build the Doubling Chain

   • Connect node 1 to v₀ (1 edge). This gives exactly 1 path from 1 to v₀.
   • For each i from 0 to 59:
     • Add an edge from vᵢ to vᵢ₊₁ (this represents "not doubling" the path count).
     • Add edges from vᵢ to uᵢ, then from uᵢ to vᵢ₊₁ (this represents "doubling" the path count—now there are 2 distinct paths from vᵢ to vᵢ₊₁).
   • After this setup, the number of paths from 1 to vᵢ is exactly 2ⁱ: each step to the next v node gives two choices, leading to 2ⁱ total paths after i steps.

3. Combine to Get x

   • Convert x to its binary representation. For every position i where the binary bit is 1 (meaning 2ⁱ contributes to x), add an edge from vᵢ to n.
   • The total number of paths from 1 to n will be the sum of all 2ⁱ for which we added edges—this sum is exactly x.

Constraint Check #

• Node Count: 1 (source) + 61 (v nodes) + 60 (u nodes) + 1 (sink) = 123, which is way below the 300 limit.
• Edge Count: 1 (1→v₀) + 60×3 (each i has 3 edges) + k (k is the number of 1s in x's binary, max 60) = 1 + 180 + 60 = 241, well under the 400 limit.
• DAG & No Self-Loops: Assign node IDs in the order we created them (1, v₀, u₀, v₁, u₁, ..., v₆₀, n). All edges go from lower-numbered nodes to higher-numbered nodes, so there can’t be cycles or self-loops.

This construction works for any valid x, and fits neatly within the given constraints.

内容的提问来源于stack exchange，提问作者Binda