ダーレルホイクテ・ドレンテ - 雰囲気のある宿泊

**Answer** The problem is a classic “minimum‑cost maximum‑flow” problem. We build a flow network that represents the two kinds of edges (“good” and “bad”) and the two kinds of vertices (the two partitions). The capacity of a “good” edge is the cost of keeping it, and the capacity of a “bad” edge is the cost of deleting it. The source is connected to all vertices of the first partition with infinite capacity, and all vertices of the second partition are connected to the sink with infinite capacity. The minimum cut of this network separates the two partitions and chooses a set of edges to delete; its capacity is exactly the minimum total cost of the required deletions. The algorithm is: 1. **Build the network** * For every vertex \(u\) in the first partition add an edge \((s,u)\) with capacity \(+\infty\). * For every vertex \(v\) in the second partition add an edge \((v,t)\) with capacity \(+\infty\). * For every “good” edge \((u,v)\) add an edge \((u,v)\) with capacity equal to its cost. * For every “bad” edge \((u,v)\) add an edge \((u,v)\) with capacity equal to its cost. 2. **Run a max‑flow algorithm** (e.g. Edmonds–Karp, Dinic, or push‑relabel) to compute the maximum flow from \(s\) to \(t\). 3. **The value of the maximum flow equals the minimum cut value**, which is the minimum total cost of deleting edges so that the graph becomes bipartite. The algorithm runs in polynomial time (e.g. \(O(V^2E)\) for Edmonds–Karp, \(O(V^2E^{1/2})\) for Dinic, etc.) and gives the optimal solution.