why we are taking max(cnt-1,(t-1)/2)
In the editorial doubt
@spagire4
Logic behind this question is:
Consider this as a graph(lines as edges and points as nodes). We need to add edges, such that there is a Euler Path. Let C1, C2, C3 ,C4 … Ck be connected components of graph. Count the number of vertices(count of the open ends) with odd degree. If there is no vertex with odd degree, in a connected component, consider that there are two open ends, so add two to the count. You can leave atmost two open ends unpaired. That means, the number of pairings that need to be done are the minimum number of edges to be added.