Yeah. We define the bipartite graph $T$ so that its vertices and edges correspond to the (good and bad) points and segments in the problem. We denote its parts by $V_1$ and $V_2.$ We need also the graph $T',$ a copy of $T.$ For each vertex $A$ of $T$ the corresponding vertex of $T'$ is denoted by $A',$ the parts of $T'$ are $V'_1$ and $V'_2.$ The union of $T$ and $T'$ is a bipartite graph; we will consider the union of $V_1$ and $V'_2$ as one part and the union of $V_2$ and $V'_1$ as another part of this graph. If a vertex $A$ in $T$ has degree $d < 100,$ the vertex $A'$ lies in the other part and has the same degree. We can add $100 - d$ edges between $A$ and $A'$ and repeat this procedure until we get a bipartite graph $Q$ such that $T$ is its subgraph and all the vertices of $Q$ have degree $100.$ The graph $Q$ is regular (that is, all its vertices have the same degree). Thus we have constructed a regular graph. This graph can have multiple edges; the reader is advised to check that our argument is not affected by it. We need the following well-known Hall’s theorem. If a bipartite graph $G$ satisfies Hall’s condition, then $G$ contains a set $R$ of edges such that each vertex of $G$ is the end of exactly one edge in $R.$ Such a set is called a perfect matching in $G.$ The Hall’s condition requires that for each $k$ and each $k$ vertices belonging to the same part of $G,$ the number of vertices adjacent to them is at least $k.$ Let us check the Hall’s condition for the graph $Q.$ If for $k$ vertices in one part there are only $l < k$ vertices adjacent to them, the edges between these vertices have exactly $100k$ ends in one part and at most $100l < 100k$ vertices in another part, a contradiction. Thus $Q$ admits a perfect matching. We can color all the edges of this matching with the first color and remove them from $Q.$ A regular graph remains with vertices of degree $99.$ In this graph we can again find a perfect matching, color its edges with the second color and remove them. Repeating this operation $100$ times (we like simple monotonous work) we color all the edges of $Q$ with $100$ colors so that all the edges beginning in each vertex have different colors. Such coloring is called edge coloring of $Q.$ Since $T$ is a subgraph of $Q,$ it also admits an edge coloring with $100$ colors. It remains to split each color $s$ into two hues $s_1$ and $s_2$ and color two halves of each edge of color s with different hues $s_1$ and $s_2.$