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##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/graph-theory/bipartite-graphs/building-teams-implementation

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░████████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/graph-theory/bipartite-graphs/building-teams-implementation

Repovive

CSES 1668 Building Teams - Implementation - Bipartite Graphs | Graph Theory | Repovive

Repovive.

Graph Theory37 sections · 1633 units81h total

Open in Course

CSES 1668 Building Teams - Implementation

The code

Here is the full solution:

function buildTeams():
    read n, m
    graph := array of n + 1 empty lists
    for i from 1 to m:
        read a, b
        graph[a].append(b)
        graph[b].append(a)
        color := array of size n + 1, all 0
        for i from 1 to n:
            if color[i] = 0 then
                if not bfs(i, 1, graph, color) then
                    print "IMPOSSIBLE"
                    return
    for i from 1 to n:
        print color[i]

The BFS function is the same as before: color nodes, check conflicts. On success, output each color plus one.

This runs in $O(V + E)$ time and uses $O(V)$ space.