python solution

# base case when building a lib is cheaper than a road
if c_lib <= c_road:
    return n*c_lib

# case when have to build roads

# initialize array to keep track of visited cities
    # 'w' not visited
    # 'g' being visited
    # 'b' finished visiting
visited = ['w'] * (n+1)

# initialize dictionary with all cities as keys
adj = {}
for city in range(1, n+1):
    adj[city] = []

# fill in the dictionary by recording adjacencies
for pair in cities:
    c1 = pair[0]
    c2 = pair[1]
    # record cities as adjacent
    adj[c1].append(c2)
    adj[c2].append(c1)


# integer list of components with ints representing their sizes
components = []

# depth first search
def dfs(city, count): 
    this_count = 1
    visited[city] = 'g'
    # go through all adjacent cities
    for adjacent in adj[city]:
        # if city has not been visited before 
        if visited[adjacent] == 'w':
            # run dfs on that city
            this_count += dfs(adjacent, count)
    # finished visiting this city
    visited[city] = 'b'
    return this_count


for city in range(1, n+1):
    if visited[city] == 'b':
        continue
    count = 0
    count = dfs(city, count)
    components.append(count)
print(components)

cost = c_lib * len(components)
for comp in components:
    cost += (comp-1)*c_road
return cost

'''