diff --git a/graph-theory/clique/solution.py b/graph-theory/clique/solution.py
new file mode 100644
index 000000000000..8396a97e2e71
--- /dev/null
+++ b/graph-theory/clique/solution.py
@@ -0,0 +1,19 @@
+# See https://en.wikipedia.org/wiki/Tur%C3%A1n_graph for the formulae we used here.
+# We first use the approximation (as given in the challege hint) to get started,
+# then we increment the result until the exact condition is met.
+
+import math
+
+
+def exact(n, k):
+    return (
+        n * n - (n % k) * math.ceil(n / k) ** 2 - (k - n % k) * math.floor(n / k) ** 2
+    ) / 2
+
+
+for case in range(int(input())):
+    N, M = map(int, input().split())
+    k = math.ceil(N * N / (N * N - 2 * M))
+    while M > exact(N, k):
+        k += 1
+    print(k)
diff --git a/graph-theory/jeanies-route/solution.py b/graph-theory/jeanies-route/solution.py
new file mode 100644
index 000000000000..262c4efe0a46
--- /dev/null
+++ b/graph-theory/jeanies-route/solution.py
@@ -0,0 +1,74 @@
+# Welp. I had tried my best, but this challenge needs so many steps
+# and thus is quite verbose to write. I will try to make myself clear.
+#
+# First we notice that the edges used are always the same no matter
+# where we start or finish. These paths form a subtree of the original
+# graph (which is the smallest subtree containing all the delivery cities).
+#
+# When we start and finish at the same city, the total distance is always
+# twice of the sum of all paths in the said subtree. And when we start
+# and finish in different cities, the distance should subtract the path
+# between these two cities. Thus we must maximize the distance between
+# the starting city and the finishing city.
+#
+# Using Dfs will greatly simplify the code. But the N seems too huge to
+# do the recursion properly. Maybe next time.
+#
+# So here's what to do:
+# 1. Find the said subtree
+# 2. Find the longest path in it
+# 3. Do the subtraction
+#
+# Time complexity: O(N)
+
+from collections import deque
+
+N, K = map(int, input().split())
+paths = [[] for i in range(N)]
+delivery = set(map(lambda x: int(x) - 1, input().split()))
+for i in range(N - 1):
+    frm, to, d = map(int, input().split())
+    paths[frm - 1].append((to - 1, d))
+    paths[to - 1].append((frm - 1, d))
+# pick one delivery city as the root and construct a tree
+parent = [-1] * N
+root = next(iter(delivery))
+q = deque([root])
+while q:
+    x = q.popleft()
+    for path in paths[x]:
+        if path[0] != parent[x]:
+            q.append(path[0])
+            parent[path[0]] = x
+# find unused cities and discard them, getting the subtree
+keep = [False] * N
+for i in delivery:
+    x = i
+    while x != -1 and not keep[x]:
+        keep[x] = True
+        x = parent[x]
+paths = [[path for path in paths[i] if keep[path[0]] and keep[i]] for i in range(N)]
+
+
+# now we find the longest path in the tree
+def furthest_vertex(
+    start,
+):  # return a tuple contains the vertex index and the path length
+    result = (start, 0)
+    visited = [False] * N
+    q = deque([result])
+    while q:  # run a Bfs from 'start' and count the path length from it
+        x = q.popleft()
+        if x[1] > result[1]:  # keep track of the furthest vertex
+            result = x
+        visited[x[0]] = True
+        for path in paths[x[0]]:
+            if not visited[path[0]]:
+                q.append((path[0], x[1] + path[1]))
+    return result
+
+
+start = next(i for i, x in enumerate(paths) if len(x))  # get the first city in list
+longest_path = furthest_vertex(furthest_vertex(start)[0])[1]
+# all paths exist in both connected cities, so no need to multiply by 2 here
+print(sum(sum(x[1] for x in l) for l in paths) - longest_path)